poj 3177 Redundant Paths 边双连通分量

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 
#include 
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 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;	
}