HDU 5765 Bonds Subtle pressure violence

Test instructions: Give a 20-point undirected graph to find out how many minimum cut sets are included in each edge

Analysis: The minimum cut set is the set of edges, it is clear that the minimum cut set happens to be divided into two parts (this if you do not understand can be used to disprove the law)

And then there's the official puzzle: http://bestcoder.hdu.edu.cn/blog/2016-school Training 4th 1003

In fact, the general idea is to enumerate a possible cut set each time, that is, the pressure enumeration, which has an illegal, can be pre-processing to mark all the legitimate state

All that's left is the code, and it's important to look at the code details.

#include <cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<map>#include<queue>#include<vector>using namespaceStd;typedefLong LongLL;Const intN = (1<< -)+5;intg[n],sum[n],t,kase,n,m,u[205],v[205],tot;BOOLCan[n];queue<int>Q;inlineintLowbit (intx) {returnx& (-x);}intMain () {scanf ("%d",&u);  while(t--) {scanf ("%d%d", &n,&m), tot=1<<N; Memset (g,0,sizeof(g)); Memset (CAN,false,sizeof(Can)); memset (SUM,0,sizeof(sum));  for(intI=0; i<m;++i) {scanf ("%d%d",&u[i],&V[i]); g[1<<u[i]]|=1<<V[i]; g[1<<v[i]]|=1<<U[i]; }     for(intI=1; i<tot;++i) g[i]|=g[i-lowbit (i)]|g[lowbit (i)];  for(intI=0; i<n;++i) Q.push (1<<i), can[1<<i]=true;  while(!Q.empty ()) {      intu=Q.front (); Q.pop (); intgo=g[u]^ (g[u]&T);  while(GO) {intTo=lowbit (GO) |u; if(!can[to]) Q.push (to), can[to]=true; Go-=lowbit (GO); }    }    intAll=0;  for(intI=1; i<tot;++i) {      intJ= (tot-1)^i; if(i<j&&can[i]&&Can[j]) {        ++sum[i];++sum[j];++All ; }    }     for(intj=0; j<n;++j) {       for(inti=tot-1;i>0;--i)if(! (i& (1&LT;&LT;J)) sum[i]+=sum[i^ (1<<j)]; } printf ("Case #%d:",++Kase);  for(intI=0; i<m;++i) printf ("%d", all-sum[(1<<u[i]) | (1<<V[i]); printf ("\ n"); }  return 0;}

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HDU 5765 Bonds Subtle pressure violence