poj 3281 最大流建图

题目链接:http://poj.org/problem?id=3281

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

#define maxn 105
#define maxe 20000
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,F,D;
    cin>>N>>F>>D;
    int s = 0, t = F+2*N+D+1;
    for(int i=1;i<=F;i++)  solver.addedge(s,i,1);
    for(int i=1;i<=N;i++)  solver.addedge(F+i,F+N+i,1);
    for(int i=1;i<=D;i++)  solver.addedge(F+2*N+i,t,1);
    for(int i=1;i<=N;i++){
        int food,drink,temp;
        cin>>food>>drink;
        for(int j=1;j<=food;j++){
            cin>>temp;
            solver.addedge(temp,F+i,1);
        }
        for(int j=1;j<=drink;j++){
            cin>>temp;
            solver.addedge(F+N+i,F+2*N+temp,1);
        }
    }
    printf("%d\n",solver.MaxFlow(s,t));
}
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posted @ 2013-08-19 10:40  等待最好的两个人  阅读(164)  评论(0编辑  收藏  举报