nyoj 谍战 (最小割最大流)

做完图论学网络流。。  不会先转了。。

转自http://blog.csdn.net/AQ14AQ1/article/details/38866407

如果能想到是最小割问题，那么建图思路便是水到渠成的事了。

添加一个源点S和汇点T；

把S与每个间谍相连，容量为无穷大；

把城市N（即飞机场的位置）与汇点T相连，容量为无穷大；

之间有道路的城市相连，容量为1，注意这里是双向的边；

建图完后，根据最大流最小割定理，那么直接求最大流即可。

#include<iostream>
using namespace std;
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<algorithm>

#define INS 1<<30
#define CLR(arr,v) memset(arr,v,sizeof(arr))

#define MaxV 3000
#define MaxE 100000

class MaxFlow{
public:
    void Clear(){
        CLR(h,-1); CLR(cnt,0); CLR(vis,0);
        flag = false;
        pos = top = head = total = maxflow = 0;
    }
    void add(int u,int v,int flow){
        num[pos] = v;
        sur[pos] = flow;
        next[pos] = h[u];
        h[u] = pos++;

        num[pos] = u;
        sur[pos] = 0;
        next[pos] = h[v];
        h[v] = pos++;
    }
    int GetMaxFlow(int s,int t){
        init(t);
        stk[top] = s;
        while(!flag){
            minres = INS;
            if(top < 0) top = 0;
            if(!dfs(stk[top],-1,t,minres)) continue;
            maxflow += minres;
            while(dis != -1){
                sur[dis] -= minres;
                sur[dis^1] += minres;
                dis = pre_e[dis];
            }
            top = 0;
        }
        return maxflow;
    }
private:
    int h[MaxV],num[MaxE],sur[MaxE],next[MaxE],gap[MaxV],cnt[MaxV],pre_e[MaxE],stk[MaxV],que[MaxV];
    int pos,top,head,total,maxflow,minres,dis;
    bool vis[MaxV],flag;
    void init(int n){
        que[total++] = n;
        vis[n] = true;
        while(head < total){
            int p = que[head++];
            if(head >= MaxV) head -= MaxV;
            for(int i = h[p]; i != -1 ;i = next[i]){
                if(!vis[ num[i] ]){
                    vis[ num[i] ] = true;
                    gap[ num[i] ] = gap[p] + 1;
                    cnt[ gap[ num[i] ] ]++;
                    que[total++] = num[i];
                    if(total >= MaxV) total -= MaxV;
                }
            }
        }
    }
    bool dfs(int p,int father,int n,int &minres){
        int m = minres;
        for(int i = h[p]; i != -1 ;i = next[i]){
            if(sur[i] > 0 && gap[p] - gap[ num[i] ] == 1){
                minres = min(minres,sur[i]);
                pre_e[i] = father;
                if(num[i] != n) stk[++top] = num[i];
                if(num[i] == n || dfs(num[i],i,n,minres)) {
                    if(num[i] == n) dis = i;
                    return true;
                }
                minres = m;
            }
        }
        cnt[ gap[p] ]--;
        cnt[ gap[p] + 1]++;
        top--;
        if(cnt[ gap[p] ] == 0) flag = true;
        gap[p] += 1;
        return false;
    }
}T;


int main()
{   
    int t;
    scanf("%d",&t);
    int cnt = 0;
    while (t--)
          {
          int n,m,p;
          scanf("%d%d%d",&n,&m,&p);
          
          int N = n + 1;
          T.Clear();
          
          int x;
          for (int i = 1; i <= p; ++ i)
               scanf("%d",&x),T.add(0,x,INS);
          
          for (int i = 1; i <= m; ++ i)
              {
              int u,v;
              scanf("%d%d",&u,&v);     
              T.add(u,v,1);
              T.add(v,u,1);
              }
          
          T.add(n,N,INS);
    
          printf("Case #%d: %d
",++cnt,T.GetMaxFlow(0,N));
          }
    return 0;
}