poj 1149 最大流

题目链接:http://poj.org/problem?id=1149

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>

#define maxn 1050
#define maxe 200000
using namespace std;

const int INF = 0x3f3f3f;

struct Edge{
    int u,v,flow,cap;
    int next;
    Edge(int u=0,int v=0,int flow=0,int cap=0,int next=0):
         u(u),v(v),flow(flow),cap(cap),next(next) { }
};


struct Dinic{
    int s,t;
    int d[maxn];
    int cur[maxn];
    bool vis[maxn];
    Edge edges[maxe];
    int head[maxn],cnt;

    void init(){
        memset(head,-1,sizeof(head));
        cnt = 0;
    }

    void addedge(int u,int v,int cap){
        edges[cnt] = Edge(u,v,0,cap,head[u]);
        head[u] = cnt++;
        edges[cnt] = Edge(v,u,0,0,head[v]);
        head[v] = cnt++;
    }

    bool bfs(){
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        Q.push(s);
        vis[s] = true;
        d[s] = 0;
        while(!Q.empty()){
            int u = Q.front();  Q.pop();
            for(int i=head[u];i!=-1;i=edges[i].next){
                Edge& e = edges[i];
                if(!vis[e.v] && e.cap > e.flow){
                    vis[e.v] = true;
                    d[e.v] = d[e.u] + 1;
                    Q.push(e.v);
                }
            }
        }
        return vis[t];
    }

    int dfs(int u,int res){
        if(u == t || res == 0)  return  res;  //res 残流大小;
        int flow = 0,f;
        for(int& i=cur[u];i!=-1;i=edges[i].next){
            Edge& e = edges[i];
            if(d[e.v] == d[e.u] + 1 && (f = dfs(e.v,min(res,e.cap-e.flow))) > 0){
                e.flow += f;
                edges[i^1].flow -= f;
                flow += f;
                res -= f;
                if(res == 0)  break;
            }
        }
        return flow;
    }

    int MaxFlow(int s_,int t_){
        s = s_;    t = t_;
        int flow  = 0;
        while(bfs()){
            for(int i=s;i<=t;i++)  cur[i] = head[i];
            flow += dfs(s,INF);
        }
        return flow;
    }
}solver;

int main()
{
    //freopen("E:\\acm\\input.txt","r",stdin);
    solver.init();
    int N,M;
    cin>>M>>N;
    int s = 0, t = N+M+1;
    for(int i=1;i<=M;i++){
        int a;
        scanf("%d",&a);
        solver.addedge(s,i,a);
    }
    vector<int> G[105];
    for(int i=1;i<=N;i++){
        int a;
        scanf("%d",&a);
        for(int j=1;j<=a;j++){
            int temp;
            scanf("%d",&temp);
            solver.addedge(temp,M+i,INF);
            for(int k=1;k<i;k++){
                bool flag = false;
                for(int m=0;m<G[k].size();m++)
                   if(G[k][m] == temp){
                       flag = true;
                       break;
                   }
                if(flag)  solver.addedge(M+k,M+i,INF);
            }
            G[i].push_back(temp);
        }
        int b;
        scanf("%d",&b);
        solver.addedge(M+i,t,b);
    }
    printf("%d\n",solver.MaxFlow(s,t));
}
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posted @ 2013-08-19 11:34  等待最好的两个人  阅读(127)  评论(0编辑  收藏  举报