题意：
跟poj 3352 一样，给一个有桥的无向图，求最少加多少条边后能变为边双连通的。
思路：
求边双连通分量，缩点后得到一颗树，求得树叶树leaf后答案为(leaf+1）/2。
代码：
//poj 3177
//sepNINE
#include <iostream>
#include <stack>
using namespace std;
const int maxN=5024;
const int maxM=20024;
struct Edge
{int v,next;
}edge[maxM];
int n,m,e,t,cnt;
int head[maxN],dfn[maxN],low[maxN],inStack[maxN],par[maxN],belong[maxN];
stack<int> mystack;
void tarjan(int h)
{inStack[h]=1;mystack.push(h); dfn[h]=low[h]=++t;for(int i=head[h];i!=-1;i=edge[i].next){int v=edge[i].v;if(!dfn[v]){par[v]=h;tarjan(v);low[h]=min(low[h],low[v]);}else if(inStack[v]&&v!=par[h]){low[h]=min(low[h],dfn[v]); } }if(low[h]==dfn[h]){int k;++cnt;do{k=mystack.top();belong[k]=cnt;mystack.pop();inStack[k]=0; }while(low[k]!=dfn[k]);}return ;
}
void solve()
{memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));memset(inStack,0,sizeof(inStack));memset(belong,0,sizeof(belong));while(!mystack.empty()) mystack.pop();int i,j;for(i=1;i<=n;++i)if(!dfn[i]){par[i]=-1;tarjan(i); }int leaf=0,d[maxN];memset(d,0,sizeof(d));for(i=1;i<=n;++i)for(j=head[i];j!=-1;j=edge[j].next){int u=belong[i],v=belong[edge[j].v];if(u!=v){++d[u];++d[v];} }for(i=1;i<=cnt;++i)if(d[i]==2)++leaf;printf("%d\n",(leaf+1)/2); return ;
}int main()
{while(scanf("%d%d",&n,&m)==2){t=0;e=0;cnt=0;memset(head,-1,sizeof(head));while(m--){int a,b,flag=0;scanf("%d%d",&a,&b);for(int i=head[a];i!=-1;i=edge[i].next)if(edge[i].v==b)flag=1;if(flag==1)continue;edge[e].v=b;edge[e].next=head[a];head[a]=e++;edge[e].v=a;edge[e].next=head[b];head[b]=e++;} solve();}return 0;
}