p2944 [USACO09MAR]地震损失2Earthquake Damage 2

传送门

分析

我们让s到1，关键点到t分别连流量为inf的边

于是我们可以考虑跑s到t的最小割

于是我们将所有点拆为两个点，关键点和1的两个点之间连inf，其余点连1

将原图的边也连上，流量为inf

于是跑最小割即可

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int dx[] = {,-,,};
const int dy[] = {,,,-};
const int inf = 1e9+;
int n,m,s,t,head[],to[],nxt[],w[],ano[],cnt,level[],cur[];
int used[];
inline void add(int x,int y,int z){
    nxt[++cnt]=head[x];
    head[x]=cnt;
    to[cnt]=y;
    w[cnt]=z;
    ano[cnt]=cnt+;
    nxt[++cnt]=head[y];
    head[y]=cnt;
    to[cnt]=x;
    w[cnt]=;
    ano[cnt]=cnt-;
}
inline bool bfs(){
    memset(level,-,sizeof(level));
    queue<int>q;
    level[s]=;
    q.push(s);
    while(!q.empty()){
      int x=q.front();
      q.pop();
      for(int i=head[x];i;i=nxt[i])
        if(level[to[i]]==-&&w[i]){
          level[to[i]]=level[x]+;
          if(to[i]==t)return ;
          q.push(to[i]);
        }
    }
    return ;
}
inline int dfs(int x,int flow){
    if(x==t||!flow)return flow;
    int res=;
    cur[x]=head[x];
    for(int i=cur[x];i;i=nxt[i]){
      cur[x]=i;
      if(level[to[i]]==level[x]+&&w[i]){
          int f=dfs(to[i],min(w[i],flow-res));
          w[i]-=f;
          res+=f;
          w[ano[i]]+=f;
      }
    }
    if(!res)level[x]=-;
    return res;
}
int main(){
    int i,j,k,Ans=;
    scanf("%d%d%d",&n,&m,&k);
    s=*n+,t=s+;
    add(s,,inf);
    add(,n+,inf);
    for(i=;i<=m;i++){
      int x,y;
      scanf("%d%d",&x,&y);
      add(x+n,y,inf);
      add(y+n,x,inf);
    }
    for(i=;i<=k;i++){
      int x;
      scanf("%d",&x);
      add(x+n,t,inf);
      add(x,x+n,inf);
      used[x]=;
    }
    for(i=;i<=n;i++){
      if(used[i])continue;
      add(i,i+n,);
    }
    while(bfs())while(int a=dfs(s,inf))Ans+=a;
    cout<<Ans;
    return ;
}