Uva 10765 - Doves and bombs 求割点..找去掉某点一个连通图会变成几个部分..

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题意:

给了一个联通无向图...现在问去掉某个点..会让图变成几个联通块?...输出的..按分出的从多到小..若相等..输出标号从小到大.输出M个...

题解:

从求割点的过程演变得出一个点去掉会将图变成几个联通块..

也就是在做tarjan时...当一个边dfn[u]<=low[v]时..其分成的块个数++..但值得注意的同样是一个联通快第一个进去的点.个数要-1...

Program:

#include<iostream>#include<stdio.h>#include<string.h>#include<set>#include <stack>#include<queue>#include<algorithm>#include<cmath>#define oo 1000000007#define ll long long#define pi acos(-1.0)#define MAXN 10005#define MAXM 500505using namespace std;     struct node{      int v,u,id,next; }edge[MAXM];struct NODE{      int id,num;}ans[MAXN];int Ne,_next[MAXN],dfn[MAXN],DfsIndex,low[MAXN]; set<int> Ans[MAXN];void addedge(int u,int v,int id){      edge[++Ne].next=_next[u],_next[u]=Ne;      edge[Ne].u=u,edge[Ne].v=v,edge[Ne].id=id;}void tarjan(int u,int id)  {        dfn[u]=low[u]=++DfsIndex;         ans[u].id=u,ans[u].num=1;      for (int k=_next[u];k;k=edge[k].next)           if (edge[k].id!=id)           {                  int v=edge[k].v;                      if (!dfn[v])                  {                         tarjan(v,edge[k].id);                         low[u]=min(low[u],low[v]);                         if (dfn[u]<=low[v]) ans[u].num++;                  }else                         low[u]=min(low[u],dfn[v]);          }   }  bool cmp(NODE a,NODE b){      if (a.num!=b.num) return a.num>b.num;      return a.id<b.id;}int main(){            int u,v,n,m,id,i;        while (scanf("%d%d",&n,&m) && n && m)      {               Ne=id=0,memset(_next,0,sizeof(_next));               while (scanf("%d%d",&u,&v) && u!=-1)                   addedge(u,v,++id),addedge(v,u,id);               memset(dfn,0,sizeof(dfn));                 DfsIndex=0;               for (u=0;u<n;u++)                  if (!dfn[u])                   {                              tarjan(u,0);                          ans[u].num--;                  }                  sort(ans,ans+n,cmp);                for (i=0;i<m;i++) printf("%d %d\n",ans[i].id,ans[i].num);                puts("");      }      return 0;}