这次就是最小流了
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const ll INF=0x3f3f3f3f3f3f3f3f; 5 const int N=50200; 6 7 struct node{ 8 ll t,cap,flow,next; //cap容量，flow流量 9 }e[N*12]; 10 int head[N],vis[N],cnt; 11 void add(int u,int v,ll cap) //u->v容量为cap 12 { 13 e[cnt]=node{v,cap,0,head[u]}; 14 head[u]=cnt++; 15 e[cnt]=node{u,0,0,head[v]}; //容量为0的反向边 16 head[v]=cnt++; 17 } 18 int d[N]; //bfs深度 19 bool bfs(int s,int t) //O(n+m) 20 { 21 memset(d,0,sizeof(d)); 22 queue<int>q; 23 q.push(s); 24 d[s]=1; 25 while(!q.empty()) 26 { 27 int u=q.front();q.pop(); 28 for(int i=head[u];~i;i=e[i].next) 29 { 30 int v=e[i].t; 31 if(d[v]==0&&e[i].cap-e[i].flow>0) 32 { 33 d[v]=d[u]+1; 34 q.push(v); 35 } 36 } 37 } 38 return d[t]>0; //存在增广路 39 } 40 ll dfs(int s,int t,ll minedge) 41 { 42 if(s==t)return minedge; 43 ll flow=0; //从当前s点流出的流量 44 for(int &i=vis[s];~i;i=e[i].next) 45 { 46 int v=e[i].t; 47 if(d[v]==d[s]+1&&e[i].cap-e[i].flow>0) //层次关系&&有剩余流量 48 { 49 ll temp=dfs(v,t,min(minedge-flow,e[i].cap-e[i].flow)); 50 e[i].flow+=temp; //流量增加 51 e[i^1].flow-=temp; //反向边流量减少 52 flow+=temp; //flow已分配的流量 53 if(flow==minedge)return flow; //已达到祖先的最大流，无法再大，剪枝 54 } 55 } 56 if(flow==0)d[s]=0; //此点已无流，标记掉 57 return flow; 58 } 59 ll dinic(int s,int t) //一定要建立反向边cap=0 60 { 61 ll maxflow=0; 62 while(bfs(s,t)) //有增广路 63 { 64 memcpy(vis,head,sizeof(head)); 65 maxflow+=dfs(s,t,INF); 66 } 67 return maxflow; 68 } 69 70 71 ll bout[N],low[N*10]; 72 int main() 73 { 74 int n,m,s,t,u,v; 75 ll b,c; 76 while(cin>>n>>m>>s>>t) 77 { 78 memset(head,-1,sizeof(head)); 79 cnt=0; 80 memset(bout,0,sizeof(bout)); 81 for(int i=0;i<m;i++) 82 { 83 scanf("%d%d%lld%lld",&u,&v,&b,&c); 84 add(u,v,c-b); 85 bout[u]+=b; 86 bout[v]-=b; 87 low[i]=b; 88 } 89 int ss=n+1,tt=n+2; 90 ll sum=0; 91 for(int i=1;i<=n;i++){ 92 if(bout[i]>0)sum+=bout[i],add(i,tt,bout[i]); 93 if(bout[i]<0)add(ss,i,-bout[i]); 94 } 95 ll res=dinic(ss,tt); 96 add(t,s,INF); 97 res+=dinic(ss,tt); 98 if(sum==res) 99 { 100 printf("%lld ",e[cnt-2].flow); 101 } 102 else printf("please go home to sleep "); 103 } 104 }