「题解」kuangbin 最小生成树

• POJ-1251 Jungle Roads (水题，%c)
• POJ-1287 Networking (水)
• POJ-2031 Building a Space Station (%f浮点数尴尬精度，两球间距离)
• POJ-2421 Constructing Roads (一些边已建好，简单处理一下)
• ZOJ-1586 QS Network (处理一下边权)
• HDU-1233 还是畅通工程 (水)
• HDU-1875 畅通工程再续 (浮点数，条件连边)
• HDU-1301 Jungle Roads (重复 -> POJ-1251)
• POJ-2349 Arctic Network (第 K 大边)
• POJ-1751 Highways (求最小生成树的边，加 0 权边)
• UVA-10147 Highways (多组的 POJ-1751) +
• POJ-1258 Agri-Net (水)
• POJ-3026 Borg Maze (BFS + Kruskal，用最短路建图，垃圾输入)
• POJ-1789 Truck History (Prim，Kruskal 超时)
• POJ-1679 The Unique MST (次小生成树)

代码最后附简单题解

模板

Kruskal

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* Kruskal 算法求 MST */#include <algorithm>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;} //排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);} //传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}

Prim

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* Prim 求 MST * 耗费矩阵 cost[][]，标号从 0 开始，0∼n-1 * 返回最小生成树的权值，返回 -1 表示原图不连通 */const int INF = 0x3f3f3f3f;const int MAXN = 110;bool vis[MAXN];int lowc[MAXN];//点是 0 n-1int Prim(int cost[][MAXN], int n) { int ans = 0; memset(vis, false, sizeof(vis)); vis[0] = true; for (int i = 1; i < n; i++) lowc[i] = cost[0][i]; for (int i = 1; i < n; i++) { int minc = INF; int p = -1; for (int j = 0; j < n; j++) if (!vis[j] && minc > lowc[j]) { minc = lowc[j]; p = j; } if (minc == INF) return -1; //原图不连通 ans += minc; vis[p] = true; for (int j = 0; j < n; j++) { if (!vis[j] && lowc[j] > cost[p][j]) lowc[j] = cost[p][j]; } } return ans;}

SecondMST (Prim)

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* 次小生成树 * 求最小生成树时，用数组 Max[i][j] 来表示 MST 中 i 到 j 最大边权 * 求完后，直接枚举所有不在 MST 中的边，替换掉最大边权的边，更新答案 * 点的编号从 0 开始 */#include <algorithm>#include <cmath>using namespace std;const int MAXN = 110;const int INF = 0x3f3f3f3f;bool vis[MAXN];int lowc[MAXN];int pre[MAXN];int Max[MAXN][MAXN];// Max[i][j] 表示在最小生成树中从 i 到 j 的路径中的最大边权bool used[MAXN][MAXN];int Prim(int cost[][MAXN], int n) { int ans = 0; memset(vis, false, sizeof(vis)); memset(Max, 0, sizeof(Max)); memset(used, false, sizeof(used)); vis[0] = true; pre[0] = -1; for (int i = 1; i < n; i++) { lowc[i] = cost[0][i]; pre[i] = 0; } lowc[0] = 0; for (int i = 1; i < n; i++) { int minc = INF; int p = -1; for (int j = 0; j < n; j++) if (!vis[j] && minc > lowc[j]) { minc = lowc[j]; p = j; } if (minc == INF) return -1; ans += minc; vis[p] = true; used[p][pre[p]] = used[pre[p]][p] = true; for (int j = 0; j < n; j++) { if (vis[j] && j != p) Max[j][p] = Max[p][j] = max(Max[j][pre[p]], lowc[p]); if (!vis[j] && lowc[j] > cost[p][j]) { lowc[j] = cost[p][j]; pre[j] = p; } } } return ans;}int smst(int cost[][MAXN], int n, int ans) { int Min = INF; for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) if (cost[i][j] != INF && !used[i][j]) { Min = min(Min, ans + cost[i][j] - Max[i][j]); } if (Min == INF) return -1; //不存在 return Min;}

代码

POJ-1251 Jungle Roads

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// POJ-1251 Jungle Roads// https://vjudge.net/problem/POJ-1251#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int n; while (~scanf("%d", &n)) { if (n == 0) break; tol = 0; for (int i = 0; i < n - 1; i++) { char x; int m; scanf(" %c%d", &x, &m); for (int j = 0; j < m; j++) { char y; int w; scanf(" %c%d", &y, &w); addedge(x - 'A', y - 'A', w); } } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 裸题，scanf %c 读入尴尬// scanf(" %c%d", &x, &m);

POJ-1287 Networking

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// POJ-1287 Networking// https://vjudge.net/problem/POJ-1287#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int n, m; while (~scanf("%d", &n)) { if (n == 0) break; tol = 0; scanf("%d", &m); for (int i = 0; i < m; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); addedge(a, b, c); } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 裸题

POJ-2031 Building a Space Station

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// POJ-2031 Building a Space Station// https://vjudge.net/problem/POJ-2031#include <string.h>#include <algorithm>#include <cmath>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct { int u, v; double w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, double w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1double Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 double ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; double w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}double len(double x, double y, double z, double xx, double yy, double zz) { return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y) + (zz - z) * (zz - z));}double cirlen(double x, double y, double z, double r, double xx, double yy, double zz, double rr) { double ll = len(x, y, z, xx, yy, zz); if (ll - r - rr < 1e-5) { return 0; } else { return ll - r - rr; }}double xp[MAXN];double yp[MAXN];double zp[MAXN];double rp[MAXN];int main() { int n; while (~scanf("%d", &n)) { if (n == 0) break; tol = 0; for (int i = 0; i < n; i++) { scanf("%lf%lf%lf%lf", &xp[i], &yp[i], &zp[i], &rp[i]); } for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { addedge(i, j, cirlen(xp[i], yp[i], zp[i], rp[i], xp[j], yp[j], zp[j], rp[j])); } } double ans = Kruskal(n); printf("%.3fn", ans); } return 0;}// 浮点数的最小生成树，球之间的距离，距离小于零，就建一条权为 0 的边// POJ尴尬精度 %f

POJ-2421 Constructing Roads

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// POJ-2421 Constructing Roads// https://vjudge.net/problem/POJ-2421#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;} //排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);} //传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int mapx[MAXN][MAXN];int main() { int N; scanf("%d", &N); for (int i = 1; i <= N; i++) { for (int j = 1; j <= N; j++) { scanf("%d", &mapx[i][j]); } } int Q; scanf("%d", &Q); for (int i = 0; i < Q; i++) { int a, b; scanf("%d%d", &a, &b); mapx[a][b] = 0; mapx[b][a] = 0; } for (int i = 1; i <= N; i++) { for (int j = i + 1; j <= N; j++) { if (i == j) continue; addedge(i, j, min(mapx[i][j], mapx[j][i])); } } int ans = Kruskal(N); printf("%dn", ans);}// 有一些路已建好，把边权改为 0 就好了

ZOJ-1586 QS Network

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// ZOJ-1586 QS Network// https://vjudge.net/problem/ZOJ-1586#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 1100; //最大点数const int MAXM = 2000000; //最大边数int F[MAXN]; //并查集使用int codec[MAXN]; // *&*struct Edge { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int T; scanf("%d", &T); while (T--) { tol = 0; int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &codec[i]); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { int w; scanf("%d", &w); if (i < j) addedge(i, j, w + codec[i] + codec[j]); } } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 加边时，加一下适配器的价格

HDU-1233 还是畅通工程

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// HDU-1233 还是畅通工程// https://vjudge.net/problem/HDU-1233#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct Edge { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;} //排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);} //传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int n; while (~scanf("%d", &n)) { if (n == 0) break; tol = 0; int m = n * (n - 1) / 2; for (int i = 0; i < m; i++) { int a, b, w; scanf("%d%d%d", &a, &b, &w); addedge(a, b, w); } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 裸题

HDU-1875 畅通工程再续

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// HDU-1875 畅通工程再续// https://vjudge.net/problem/HDU-1875#include <string.h>#include <algorithm>#include <cmath>#include <cstdio>using namespace std;const int MAXN = 310; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct Edge { int u, v; double w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, double w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1double Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 double ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; double w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}double len(double x, double y, double xx, double yy) { return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y));}double xp[MAXN];double yp[MAXN];int main() { int T; scanf("%d", &T); while (T--) { int n; scanf("%d", &n); if (n == 0) break; tol = 0; for (int i = 1; i <= n; i++) { scanf("%lf%lf", &xp[i], &yp[i]); } for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { double xly = len(xp[i], yp[i], xp[j], yp[j]); if (xly >= 10 && xly <= 1000) addedge(i, j, xly * 100); } } double ans = Kruskal(n); if (ans > 0) { printf("%.1fn", ans); } else { printf("oh!n"); } } return 0;}// 浮点数，条件连边

HDU-1301 Jungle Roads

123456 大专栏  「题解」kuangbin 最小生成树789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778

// HDU-1301 Jungle Roads// https://vjudge.net/problem/HDU-1301#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct Edge { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int n; while (~scanf("%d", &n)) { if (n == 0) break; tol = 0; for (int i = 0; i < n - 1; i++) { char x; int m; scanf(" %c%d", &x, &m); for (int j = 0; j < m; j++) { char y; int w; scanf(" %c%d", &y, &w); addedge(x - 'A', y - 'A', w); } } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 竟然有重复的题，一模一样 -> POJ-1251// 裸题，scanf %c 读入尴尬// scanf(" %c%d", &x, &m);

POJ-2349 Arctic Network

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// POJ-2349 Arctic Network// https://vjudge.net/problem/POJ-2349#include <string.h>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>using namespace std;const int MAXN = 510; //最大点数const int MAXM = 250000; //最大边数int F[MAXN]; //并查集使用vector<double> vd;struct Edge { int u, v; double w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, double w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1double Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 double ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; double w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans = max(ans, w); vd.push_back(w); // ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}double len(double x, double y, double xx, double yy) { return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y));}double xp[MAXN];double yp[MAXN];int main() { int T; scanf("%d", &T); while (T--) { vd.clear(); int n, s; scanf("%d%d", &s, &n); if (n == 0) break; tol = 0; for (int i = 1; i <= n; i++) { scanf("%lf%lf", &xp[i], &yp[i]); } for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { double xly = len(xp[i], yp[i], xp[j], yp[j]); addedge(i, j, xly); } } double ans = Kruskal(n); ans = vd[vd.size() - s]; printf("%.2fn", ans); } return 0;}// 最小生成树的第 K 大边

POJ-1751 Highways

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// POJ-1751 Highways// https://vjudge.net/problem/POJ-1751#include <string.h>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>using namespace std;const int MAXN = 1000; //最大点数const int MAXM = 1000000; //最大边数vector<int> vx;vector<int> vy;int F[MAXN]; //并查集使用struct Edge { int u, v; double w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, double w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1double Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 double ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; double w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; if (w > 1e-5) { vx.push_back(u); vy.push_back(v); } F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}double len(double x, double y, double xx, double yy) { return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y));}double xp[MAXN];double yp[MAXN];int main() { int n; scanf("%d", &n); tol = 0; for (int i = 1; i <= n; i++) { scanf("%lf%lf", &xp[i], &yp[i]); } for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { double xly = len(xp[i], yp[i], xp[j], yp[j]); addedge(i, j, xly * 100); } } int q; scanf("%d", &q); for (int i = 0; i < q; i++) { int a, b; scanf("%d%d", &a, &b); addedge(a, b, 0); } double ans = Kruskal(n); for (int i = 0; i < vx.size(); i++) { printf("%d %dn", vx[i], vy[i]); } return 0;}// 求最小生成树的边，加 0 权边

UVA-10147 Highways

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// UVA-10147 Highways// https://vjudge.net/problem/UVA-10147#include <string.h>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>using namespace std;const int MAXN = 1000; //最大点数const int MAXM = 1000000; //最大边数vector<int> vx;vector<int> vy;int F[MAXN]; //并查集使用struct Edge { int u, v; double w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, double w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;}//排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);}//传入点数，返回最小生成树的权值，如果不连通返回 -1double Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 double ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; double w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; if (w > 0) { vx.push_back(u); vy.push_back(v); } F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}double len(double x, double y, double xx, double yy) { return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y));}double xp[MAXN];double yp[MAXN];int main() { int T; scanf("%d", &T); while (T--) { vx.clear(); vy.clear(); int n; scanf("%d", &n); tol = 0; for (int i = 1; i <= n; i++) { scanf("%lf%lf", &xp[i], &yp[i]); } for (int i = 1; i <= n; i++) { for (int j = i + 1; j <= n; j++) { double xly = len(xp[i], yp[i], xp[j], yp[j]); addedge(i, j, xly * 100); } } int q; scanf("%d", &q); for (int i = 0; i < q; i++) { int a, b; scanf("%d%d", &a, &b); addedge(a, b, 0); } double ans = Kruskal(n); for (int i = 0; i < vx.size(); i++) { printf("%d %dn", vx[i], vy[i]); } if (vx.size() == 0) printf("No new highways needn"); if (T != 0) printf("n"); } return 0;}// 求最小生成树的边，加 0 权边

POJ-1258 Agri-Net

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// POJ-1258 Agri-Net// https://vjudge.net/problem/POJ-1258#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 110; //最大点数const int MAXM = 10000; //最大边数int F[MAXN]; //并查集使用struct Edge { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;} //排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);} //传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int main() { int n; while (~scanf("%d", &n)) { tol = 0; for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { int w; scanf("%d", &w); addedge(i, j, w); } } int ans = Kruskal(n); printf("%dn", ans); } return 0;}// 裸题

POJ-3026 Borg Maze

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// POJ-3026 Borg Maze// https://vjudge.net/problem/POJ-3026#include <string.h>#include <algorithm>#include <cstdio>#include <queue>using namespace std;const int MAXN = 1100; //最大点数const int MAXM = 1000000; //最大边数int F[MAXN]; //并查集使用struct Edge { int u, v, w;} edge[MAXM]; //存储边的信息，包括起点/终点/权值int tol; //边数，加边前赋值为 0void addedge(int u, int v, int w) { edge[tol].u = u; edge[tol].v = v; edge[tol++].w = w;} //排序函数，讲边按照权值从小到大排序bool cmp(Edge a, Edge b) { return a.w < b.w; }int find(int x) { if (F[x] == -1) return x; else return F[x] = find(F[x]);} //传入点数，返回最小生成树的权值，如果不连通返回 -1int Kruskal(int n) { memset(F, -1, sizeof(F)); sort(edge, edge + tol, cmp); int cnt = 0; //计算加入的边数 int ans = 0; for (int i = 0; i < tol; i++) { int u = edge[i].u; int v = edge[i].v; int w = edge[i].w; int t1 = find(u); int t2 = find(v); if (t1 != t2) { ans += w; F[t1] = t2; cnt++; } if (cnt == n - 1) break; } if (cnt < n - 1) return -1; //不连通 else return ans;}int mapx[MAXN][MAXN];int dirx[10] = {0, 1, 0, -1};int diry[10] = {1, 0, -1, 0};int n, m;void bfs(int x, int y) { bool vis[MAXN][MAXN]; memset(vis, false, sizeof(vis)); queue<pair<pair<int, int>, int> > Q; Q.push(make_pair(make_pair(x, y), 0)); vis[x][y] = true; while (!Q.empty()) { int xt = Q.front().first.first; int yt = Q.front().first.second; int times = Q.front().second; Q.pop(); if (mapx[x][y] != mapx[xt][yt] && mapx[xt][yt] != 0) { addedge(mapx[x][y], mapx[xt][yt], times); } for (int i = 0; i < 4; i++) { int tx = xt + dirx[i]; int ty = yt + diry[i]; if (tx >= 0 && ty >= 0 && tx < n && ty < m && !vis[tx][ty]) { if (mapx[tx][ty] != -1) { Q.push(make_pair(make_pair(tx, ty), times + 1)); vis[tx][ty] = true; } } } }}int main() { int T; scanf("%d", &T); char str[100]; while (T--) { tol = 0; memset(mapx, -1, sizeof(mapx)); int cnt = 0; scanf("%d%d", &n, &m); gets(str); for (int i = 0; i < m; i++) { gets(str); for (int j = 0; j < n; j++) { if (str[j] == 'A' || str[j] == 'S') { mapx[i][j] = ++cnt; } else if (str[j] == ' ') { mapx[i][j] = 0; } } } for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (mapx[i][j] > 0) { bfs(i, j); } } } int ans = Kruskal(cnt); printf("%dn", ans); } return 0;}// BFS + Kruskal// 用无权图最短路权值加边// 一开始 MAXN 和 MAXM 开小了，wa...wa...wa...

POJ-1789 Truck History

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// POJ-1789 Truck History// https://vjudge.net/problem/POJ-1789#include <string.h>#include <algorithm>#include <cstdio>using namespace std;const int MAXN = 2010; //最大点数const int INF = 0x3f3f3f3f;bool vis[MAXN];int lowc[MAXN];int cost[MAXN][MAXN];//点是 0 n-1int Prim(int cost[][MAXN], int n) { int ans = 0; memset(vis, false, sizeof(vis)); vis[0] = true; for (int i = 1; i < n; i++) lowc[i] = cost[0][i]; for (int i = 1; i < n; i++) { int minc = INF; int p = -1; for (int j = 0; j < n; j++) if (!vis[j] && minc > lowc[j]) { minc = lowc[j]; p = j; } if (minc == INF) return -1; //原图不连通 ans += minc; vis[p] = true; for (int j = 0; j < n; j++) { if (!vis[j] && lowc[j] > cost[p][j]) lowc[j] = cost[p][j]; } } return ans;}char str[MAXN][10];int len(int a, int b) { int ans = 0; for (int i = 0; i < 7; i++) { if (str[a][i] != str[b][i]) { ans++; } } return ans;}int main() { int n; while (~scanf("%d", &n)) { if (n == 0) break; for (int i = 0; i < n; i++) { scanf("%s", str[i]); } for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (i == j) continue; cost[i][j] = len(i, j); } } int ans = Prim(cost, n); printf("The highest possible quality is 1/%d.n", ans); } return 0;}// Kruskal 超时，用 Prim !

POJ-1679 The Unique MST

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// POJ-1679 The Unique MST// https://vjudge.net/problem/POJ-1679#include <string.h>#include <algorithm>#include <cstdio>#include <cmath>using namespace std;const int MAXN = 110;const int INF = 0x3f3f3f3f;bool vis[MAXN];int lowc[MAXN];int pre[MAXN];int Max[MAXN][MAXN];// Max[i][j] 表示在最小生成树中从 i 到 j 的路径中的最大边权bool used[MAXN][MAXN];int Prim(int cost[][MAXN], int n) { int ans = 0; memset(vis, false, sizeof(vis)); memset(Max, 0, sizeof(Max)); memset(used, false, sizeof(used)); vis[0] = true; pre[0] = -1; for (int i = 1; i < n; i++) { lowc[i] = cost[0][i]; pre[i] = 0; } lowc[0] = 0; for (int i = 1; i < n; i++) { int minc = INF; int p = -1; for (int j = 0; j < n; j++) if (!vis[j] && minc > lowc[j]) { minc = lowc[j]; p = j; } if (minc == INF) return -1; ans += minc; vis[p] = true; used[p][pre[p]] = used[pre[p]][p] = true; for (int j = 0; j < n; j++) { if (vis[j] && j != p) Max[j][p] = Max[p][j] = max(Max[j][pre[p]], lowc[p]); if (!vis[j] && lowc[j] > cost[p][j]) { lowc[j] = cost[p][j]; pre[j] = p; } } } return ans;}int smst(int cost[][MAXN],int n,int ans){ int Min=INF; for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) if(cost[i][j]!=INF && !used[i][j]) { Min=min(Min,ans+cost[i][j]-Max[i][j]); } if(Min==INF)return -1;//不存在 return Min;}int costx[MAXN][MAXN];int main() { int T; scanf("%d", &T); while (T--) { memset(costx, 0x3f, sizeof(costx)); int n, m; scanf("%d%d", &n, &m); int sum = 0; for (int i = 0; i < n; i++) costx[i][i] = 0; for (int i = 0; i < m; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); a--; b--; costx[a][b] = c; costx[b][a] = c; } int ans = Prim(costx, n); int ansMST = smst(costx, n,ans); if (ans == ansMST) { printf("Not Unique!n"); } else { printf("%dn", ans); } } return 0;}// 次小生成树，不知道为什么最小生成树计数不行// kuangbin 题解：https://www.cnblogs.com/kuangbin/p/3147329.html

来源：https://www.cnblogs.com/wangziqiang123/p/11718014.html