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题意：中文
思路：就是个最大权闭合图，最大利润就是总利润减去最大流
#include <queue> #include <vector> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int inf=0x3f3f3f3f; const ll INF=0x3f3f3f3f3f3f3f3fll; const int maxn=100010; struct edge{ int to,cap,rev; edge(int a,int b,int c){to=a;cap=b;rev=c;} }; vector<edge>G[maxn]; int level[maxn],iter[maxn]; void add_edge(int from,int to,int cap){ G[from].push_back(edge(to,cap,G[to].size())); G[to].push_back(edge(from,0,G[from].size()-1)); } void bfs(int s){ memset(level,-1,sizeof(level)); queue<int>que;level[s]=0; que.push(s); while(!que.empty()){ int v=que.front();que.pop(); for(unsigned int i=0;i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&level[e.to]<0){ level[e.to]=level[v]+1; que.push(e.to); } } } } int dfs(int v,int t,int f){ if(v==t) return f; for(int &i=iter[v];i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&level[v]<level[e.to]){ int d=dfs(e.to,t,min(f,e.cap)); if(d>0){ e.cap-=d; G[e.to][e.rev].cap+=d; return d; } } } return 0; } int max_flow(int s,int t){ int flow=0; while(1){ bfs(s); if(level[t]<0) return flow; memset(iter,0,sizeof(iter)); int f; while((f=dfs(s,t,inf))>0) flow+=f; } } int A[maxn]; int main(){ int n,m,u,v,val,sum=0; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) scanf("%d",&A[i]),add_edge(0,i,A[i]); for(int i=1;i<=m;i++){ scanf("%d%d%d",&u,&v,&val); sum+=val; add_edge(u,i+n,inf);add_edge(v,i+n,inf); add_edge(i+n,n+m+1,val); } int ans=max_flow(0,n+m+1); printf("%d\n",sum-ans); return 0; }