题目略,就是求最小割
那么 最小割=最大流
这里要求点权和最小,可以通过拆点转化为边权
#include<iostream> #include<algorithm> #include<cstring> #include<queue> #define IOS std::ios::sync_with_stdio(0) using namespace std; const int N=1e4,M=5e5; #define inf 1e9 int n,m; int all=1,hd[N],go[M],w[M],nxt[M]; int S,T; int dis[M],ans=0,now[M]; void add_(int x,int y,int z){ nxt[++all]=hd[x]; hd[x]=all; go[all]=y; w[all]=z; swap(x,y); nxt[++all]=hd[x]; hd[x]=all; go[all]=y; w[all]=0; } bool bfs(){ for(int i=0;i<M;i++)dis[i]=inf; queue<int> q; q.push(S); now[S]=hd[S]; dis[S]=0; while(q.empty()==0){ int x=q.front(); q.pop(); for(int i=hd[x];i;i=nxt[i]){ int y=go[i]; if(w[i]>0&&dis[y]==inf){ dis[y]=dis[x]+1; now[y]=hd[y]; q.push(y); if(y==T) return 1; } } } return 0; } int dfs(int x,int sum){ if(x==T) return sum; int k,res=0; for(int i=now[x];i&& sum ;i=nxt[i]){ now[x]=i; int y=go[i]; if(w[i]>0&&(dis[y]==dis[x]+1)){ k=dfs(y,min(sum,w[i])); if(k==0) dis[y]=inf; w[i]-=k; w[i^1]+=k; res+=k; sum-=k; } } return res; } void dinic(){ int ans =0; while(bfs()) ans+=dfs(S,inf); cout<<ans<<endl; } signed main(){ int i,x,y; cin>>n>>m>>S>>T; S+=n; for(i=1;i<=n;i++) add_(i,i+n,1); for(i=1;i<=m;i++){ cin>>x>>y,add_(x+n,y,inf); add_(y+n,x,inf); } dinic(); }