简单的费用流问题,每个人对每个任务连边,每个任务对汇点连,源点对每个人连,最大费用取反即可
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; const int maxm = 2e4+5; const int INF = 0x3f3f3f3f; struct edge{ int u, v, cap, flow, cost, nex; } edges[maxm]; int head[maxm], cur[maxm], cnt, fa[333], d[333], n, val[105][105]; bool inq[333]; void init() { memset(head, -1, sizeof(head)); } void add(int u, int v, int cap, int cost) { edges[cnt] = edge{u, v, cap, 0, cost, head[u]}; head[u] = cnt++; } void addedge(int u, int v, int cap, int cost) { add(u, v, cap, cost), add(v, u, 0, -cost); } bool spfa(int s, int t, int &flow, LL &cost) { for(int i = 0; i <= 2*n+3; ++i) d[i] = INF; //init() memset(inq, false, sizeof(inq)); d[s] = 0, inq[s] = true; fa[s] = -1, cur[s] = INF; queue<int> q; q.push(s); while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = false; for(int i = head[u]; i != -1; i = edges[i].nex) { edge& now = edges[i]; int v = now.v; if(now.cap > now.flow && d[v] > d[u] + now.cost) { d[v] = d[u] + now.cost; fa[v] = i; cur[v] = min(cur[u], now.cap - now.flow); if(!inq[v]) {q.push(v); inq[v] = true;} } } } if(d[t] == INF) return false; flow += cur[t]; cost += 1LL*d[t]*cur[t]; for(int u = t; u != s; u = edges[fa[u]].u) { edges[fa[u]].flow += cur[t]; edges[fa[u]^1].flow -= cur[t]; } return true; } int MincostMaxflow(int s, int t, LL &cost) { cost = 0; int flow = 0; while(spfa(s, t, flow, cost)); return flow; } void build_graph(int f) { init(); int s = 0, t = 2*n+1; for(int i = 1; i <= n; ++i) { addedge(s, i, 1, 0); for(int j = 1; j <= n; ++j) { addedge(i, j+n, 1, f*val[i][j]); } } for(int i = 1; i <= n; ++i) addedge(i+n, t, 1, 0); } void run_case() { cin >> n; int s = 0, t = 2*n+1; for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) cin >> val[i][j]; build_graph(1); LL cost = 0; MincostMaxflow(s, t, cost); cout << cost << "\n"; build_graph(-1); cost = 0; MincostMaxflow(s, t, cost); cout << -cost; } int main() { ios::sync_with_stdio(false), cin.tie(0); run_case(); cout.flush(); return 0; }
来源:https://www.cnblogs.com/GRedComeT/p/12298607.html