POJ 1966 - Cable TV Network 暴力枚举+最小点割

来源：互联网发布：源码加密php 编辑：程序博客网时间：2024/05/16 01:27
题意:
给出一个无向图,问最少去掉多少个点使得剩下的点不连通
题解:
暴力枚举起点和终点..跑拆点构图跑最大流求出最小点割,其中的最小值就是答案...
Program:
#include<iostream>      #include<algorithm>      #include<stdio.h>      #include<string.h>    #include<time.h>   #include<map>   #include<math.h>      #include<queue>      #define MAXN 505   #define MAXM 50005  #define oo 1000000007      #define ll long long      using namespace std;     struct Dinic                {                       struct node                     {                            int c,u,v,next;                     }edge[MAXM];                     int ne,head[MAXN];                     int cur[MAXN], ps[MAXN], dep[MAXN];                   void initial()                     {                           ne=2;                           memset(head,0,sizeof(head));                      }                     void addedge(int u, int v,int c)                     {                            edge[ne].u=u,edge[ne].v=v,edge[ne].c=c,edge[ne].next=head[u];                           head[u]=ne++;                           edge[ne].u=v,edge[ne].v=u,edge[ne].c=0,edge[ne].next=head[v];                           head[v]=ne++;                     }                     int MaxFlow(int s,int t)                     {                                                int tr, res = 0;                           int i,j,k,f,r,top;                           while(1)                           {                                  memset(dep, -1, sizeof(dep));                                  for(f=dep[ps[0]=s]=0,r=1;f!= r;)                                     for(i=ps[f++],j=head[i];j;j=edge[j].next)                                       if(edge[j].c&&dep[k=edge[j].v]==-1)                                       {                                             dep[k]=dep[i]+1;                                             ps[r++]=k;                                             if(k == t){  f=r; break;  }                                       }                                  if(dep[t]==-1) break;                                  memcpy(cur,head,sizeof(cur));                                  i=s,top=0;                                  while(1)                                  {                                       if(i==t)                                       {                                             for(tr=oo,k=0;k<top;k++)                                                if(edge[ps[k]].c<tr)                                                   tr=edge[ps[f=k]].c;                                             for(k=0;k<top;k++)                                             {                                                   edge[ps[k]].c-=tr;                                                   edge[ps[k]^1].c+=tr;                                             }                                             i=edge[ps[top=f]].u;                                             res+= tr;                                       }                                       for(j=cur[i];cur[i];j=cur[i]=edge[cur[i]].next)                                            if(edge[j].c && dep[i]+1==dep[edge[j].v]) break;                                        if(cur[i])  ps[top++]=cur[i],i=edge[cur[i]].v;                                        else                                       {                                               if(!top) break;                                               dep[i]=-1;                                               i=edge[ps[--top]].u;                                       }                                 }                           }                           return res;                    }         }T;          char c;int d[505][2];int main()     {              int N,M,i,u,v,ans;       while (~scanf("%d%d",&N,&M))      {                for (i=1;i<=M;i++)               {                       do { c=getchar(); }while (c!='(');                       scanf("%d",&d[i][0]);                       do { c=getchar(); }while (c!=',');                       scanf("%d",&d[i][1]);                       do { c=getchar(); }while (c!=')');                }               ans=oo;                for (u=0;u<N;u++)                  for (v=u+1;v<N;v++)                  {                        T.initial();                        for (i=0;i<N;i++) T.addedge(i<<1,i<<1|1,1);                        for (i=1;i<=M;i++) T.addedge(d[i][0]<<1|1,d[i][1]<<1,oo),                                           T.addedge(d[i][1]<<1|1,d[i][0]<<1,oo);                        ans=min(ans,T.MaxFlow(u<<1|1,v<<1));                  }               if (ans==oo) ans=N;               printf("%d\n",ans);      }      return 0;    }