構圖思路:

1. 源點S與頂點v連邊，容量為A

2. 頂點v與匯點T連邊，容量為B

3. 邊(a,b,c),則頂點a與頂點b連雙向邊，容量為c

則最小花費為該圖最小割即最大流。

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若兩個作業分別在不同機器運行，則之間若有邊，則必定是滿流，否則必定還有增廣路。

#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std;const int inf = 0x3f3f3f3f; const int MAXN = (int)2e5+10; int n, m; int S, T, N; int head[MAXN], idx; struct Edge{int v, f, nxt; }edge[MAXN<<3];void AddEdge(int u,int v,int f){edge[idx].v = v, edge[idx].f = f;edge[idx].nxt = head[u], head[u] = idx++;edge[idx].v = u, edge[idx].f = 0;edge[idx].nxt = head[v], head[v] = idx++; }int vh[MAXN], h[MAXN]; int dfs(int u,int flow){if(u == T) return flow;int tmp = h[u]+1, sum = flow;for(int i = head[u]; ~i; i = edge[i].nxt ){if( edge[i].f && (h[edge[i].v]+1 == h[u]) ){int p = dfs( edge[i].v, min(sum,edge[i].f));edge[i].f -= p, edge[i^1].f += p, sum -= p;if( sum == 0 || h[S] == N ) return flow - sum;}}for(int i = head[u]; ~i; i = edge[i].nxt )if( edge[i].f ) tmp = min( tmp, h[ edge[i].v ] );if( --vh[ h[u] ] == 0 ) h[S] = N;else ++vh[ h[u]=tmp+1 ];return flow-sum; } int sap(){int maxflow = 0;memset(h,0,sizeof(h));memset(vh,0,sizeof(vh));vh[0] = N;while( h[S]<N ) maxflow += dfs(S,inf);return maxflow; }int main(){while( scanf("%d%d",&n,&m) != EOF ){memset(head,-1,sizeof(head));S = 0, T = n+1, N = n+2; idx = 0;for(int i = 1; i <= n; i++){int a, b;scanf("%d%d",&a,&b);AddEdge( S, i, a );AddEdge( i, T, b );}for(int i = 0; i < m; i++){int a, b, c;scanf("%d%d%d",&a,&b,&c);AddEdge( a, b, c );AddEdge( b, a, c );}int ans = sap();printf("%d\n", ans );}return 0; } View Code

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轉載于:https://www.cnblogs.com/yefeng1627/p/3176263.html

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