HDU4289 Control 最大流

Posted 2020-07-07 shuguangzw

经典题，求去掉若干个点，使得两个点不在连通，总价值最少

所以拆点最小割，除了拆点边，流量都为无穷，拆点边是流量为价值

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <utility>
using namespace std;
typedef long long LL;
const int maxn=4e2+5;
const int INF=0x3f3f3f3f;
struct Edge
{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int d):from(u),to(v),cap(c),flow(d) {}
};
struct dinic
{
    int s,t;
    vector<Edge>edges;
    vector<int>G[maxn];
    int d[maxn];
    int cur[maxn];
    bool vis[maxn];
    void init(){
        edges.clear();
        for(int i=0;i<maxn;++i)
          G[i].clear();  
    }
    bool bfs()
    {
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty())
        {
            int x=q.front();
            q.pop();
            for(int i=0; i<G[x].size(); i++)
            {
                Edge &e= edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow)
                {
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int dfs(int x,int a)
    {
        if(x==t||a==0)return a;
        int flow=0,f;
        for(int &i=cur[x]; i<G[x].size(); i++)
        {
            Edge &e=edges[G[x][i]];
            if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow))))
            {
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return flow;
    }
    int maxflow(int s,int t)
    {
        this->s=s;
        this->t=t;
        int flow=0;
        while(bfs())
        {
            memset(cur,0,sizeof(cur));
            flow+=dfs(s,INF);
        }
        return flow;
    }
    void addedge(int u,int v,int c)
    {
        Edge x(u,v,c,0),y(v,u,0,0);
        edges.push_back(x);
        edges.push_back(y);
        int l=edges.size();
        G[u].push_back(l-2);
        G[v].push_back(l-1);
    }
}solve;
int a[maxn/2];
int main()
{
    int n,m,s,t;
    while(~scanf("%d%d",&n,&m)){
       scanf("%d%d",&s,&t);
       for(int i=1;i<=n;++i)
        scanf("%d",&a[i]);
       solve.init();
       for(int i=1;i<=m;++i){
         int u,v;
         scanf("%d%d",&u,&v);
         solve.addedge(u+n,v,INF);
         solve.addedge(v+n,u,INF);
       }
       for(int i=1;i<=n;++i)
       solve.addedge(i,i+n,a[i]);
       printf("%d\\n",solve.maxflow(s,t+n));
    }
    return 0;
}

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