题意
Sol
自己yy着写了一下Boruvka算法。
算法思想很简单,就是每次贪心的用两个联通块之间最小的边去合并。
复杂度\(O(n \log n)\),然鹅没有Kruskal跑的快,但是好像在一类生成树问题上很有用
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define fi first
#define se second
#define pb push_back
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 5001;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M;
namespace DSU {
int fa[MAXN], siz[MAXN];
void init(int N) {
for(int i = 1; i <= N; i++) fa[i] = i, siz[i] = 1;
}
int find(int x) {
return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
void unionn(int x, int y) {
int fx = find(x), fy = find(y);
if(siz[fx] > siz[fy]) swap(fx, fy);
fa[fx] = fy; siz[fy] += siz[fx];
}
};
using namespace DSU;
vector<Pair> v[MAXN];
int link[MAXN], val[MAXN];
void Boruvka() {
init(N); int ans = 0;
bool flag;
do {
flag = 0;
memset(link, -1, sizeof(link));
memset(val, 0x3f, sizeof(val));
for(int x = 1; x <= N; x++) {
int fx = find(x);
for(auto &tmp : v[x]) {
int to = tmp.fi, w = tmp.se, fy = find(to);
if(fx == fy || (w > val[fx])) continue;
link[fx] = fy; val[fx] = w;
}
}
for(int x = 1; x <= N; x++) {
int fx = find(x);
if((~link[fx]) && find(fx) != find(link[fx]))
unionn(fx, link[fx]), ans += val[fx], flag = 1;
}
}while(flag);
int f1 = find(1);
for(int i = 2; i <= N; i++) if(find(i) != f1) return (void) puts("orz");
cout << ans;
}
signed main() {
N = read(); M = read();
for(int i = 1; i <= M; i++) {
int x = read(), y = read(), w = read();
v[x].push_back({y, w});
v[y].push_back({x, w});
}
Boruvka();
}
来源:oschina
链接:https://my.oschina.net/u/4328465/blog/3274727