nyoj677 谍战

本题能够说是最小割入门级题目。

假设能想到是最小割问题。那么建图思路便是水到渠成的事了。

加入一个源点S和汇点T；

把S与每一个间谍相连。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，依据最大流最小割定理。那么直接求最大流就可以。

闲话少说，上代码:

#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;
}