POJ 1815 - Friendship - [拆点最大流求最小点割集][暴力枚举求升序割点] - [Dinic算法模板 - 邻接矩阵型]

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<queue>
  4 #define in(x) x
  5 #define out(x) x+n
  6 #define MAX 500
  7 #define INF 0x3f3f3f3f
  8 using namespace std;
  9 struct Dinic{
 10     int s,t,nv;//源点、汇点、点总数 
 11     int c[MAX][MAX],f[MAX][MAX],lev[MAX];
 12     bool vis[MAX];
 13     void addedge(int from,int to,int cap)
 14     {
 15         c[from][to]=cap, f[from][to]=0;
 16         c[to][from]=0, f[to][from]=0;
 17     }
 18     bool bfs()
 19     {
 20         memset(vis,0,sizeof(vis));
 21         queue<int> q;
 22         q.push(s);
 23         vis[s]=1;
 24         lev[s]=0;
 25         while(!q.empty())
 26         {
 27             int u=q.front();q.pop();
 28             for(int v=1;v<=nv;v++)
 29             {
 30                 if(!vis[v] && c[u][v]>f[u][v])//属于残存网络的边 
 31                 {
 32                        lev[v]=lev[u]+1;
 33                     q.push(v);
 34                     vis[v]=1;
 35                 }
 36             }
 37             
 38         }
 39         return vis[t];
 40     }
 41     int dfs(int u,int aug)
 42     {
 43         if(u==t) return aug;
 44         int res=aug,tmp;
 45         for(int v=1;v<=nv;v++)
 46         {
 47             if(lev[v]==lev[u]+1 && c[u][v]>f[u][v])
 48             {
 49                 tmp=dfs(v,min(aug,c[u][v]-f[u][v]));
 50                 f[u][v]+=tmp;
 51                 f[v][u]-=tmp;
 52                 aug-=tmp;
 53             }
 54         }
 55         return res-aug;
 56     }
 57     int maxflow()
 58     {
 59         int res=0;
 60         while(bfs()) res+=dfs(s,INF);
 61         return res;
 62     }
 63 }dinic;
 64 
 65 int n,S,T;
 66 int main(){
 67     int a;
 68     scanf("%d%d%d",&n,&S,&T);
 69     dinic.nv=n*2, dinic.s=out(S), dinic.t=in(T);
 70     for(int i=1;i<=n;++i)
 71     {
 72         if(i!=dinic.s && i!=dinic.t) dinic.addedge(in(i),out(i),1);
 73         for(int j=1,tmp;j<=n;j++)
 74         {
 75             scanf("%d",&tmp);
 76             if(i!=j && tmp) dinic.addedge(out(i),in(j),INF);
 77         }
 78     }
 79     if(dinic.c[dinic.s][dinic.t]){
 80         puts("NO ANSWER!\n");
 81         return 0;
 82     }
 83     int ans=dinic.maxflow();
 84     printf("%d\n",ans);
 85     
 86     for(int i=1;i<=n && ans;i++)
 87     {
 88         if(i==dinic.s|| i==dinic.t || !dinic.f[in(i)][out(i)]) continue;
 89         memset(dinic.f,0,sizeof(dinic.f));
 90         dinic.c[in(i)][out(i)]=0;
 91         if(dinic.maxflow()<ans)
 92         {
 93             ans--;
 94             printf("%d ",i);
 95         }
 96         else dinic.c[in(i)][out(i)]=1;
 97     }
 98     printf("\n");
 99     return 0;
100 }