nyoj677 谍战

来源：互联网发布：冬天护手霜推荐知乎编辑：程序博客网时间：2024/05/21 10:16

本题可以说是最小割入门级题目。

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。

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 100000class 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\n",++cnt,T.GetMaxFlow(0,N));          }    return 0;}

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