poj 3177 Redundant Paths【求最少添加多少条边可以使图变成双连通图】【缩点后求入度为1的点个数】

#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<algorithm>
#include<vector>
#define MAX 20010
#define INF 0x7fffff
using namespace std;
struct node
{
	int beg,end,next;
}edge[MAX];
int head[MAX],ans,bridge;
int low[MAX],dfn[MAX],in[MAX];
int dfsclock,ebccnt;
int instack[MAX],ebcno[MAX];
vector<int>newmap[MAX];
stack<int>s;
int n,m;
void init()
{
	ans=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
	edge[ans].beg=u;
	edge[ans].end=v;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void getmap()
{
	int a,b;
	while(m--)
	{
		scanf("%d%d",&a,&b);
		add(a,b);
		add(b,a);
	}
}
void tarjan(int u,int fa)
{
	int v;
	instack[u]=1;
	s.push(u);
	low[u]=dfn[u]=++dfsclock;
	bool flag=true;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(flag&&v==fa)
		{
			flag=false;
			continue;
		}
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(dfn[u]<low[v])
			    bridge++;
		}
		else if(instack[v])
		    low[u]=min(low[u],dfn[v]);
	}
	if(dfn[u]==low[u])
	{
		ebccnt++;
		while(1)
		{
			v=s.top();
			s.pop();
			instack[v]=0;
			ebcno[v]=ebccnt;
			if(v==u)
			    break;
		}
	}
}
void find()
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	memset(instack,0,sizeof(instack));
	memset(ebcno,0,sizeof(ebcno));
	dfsclock=ebccnt=bridge=0;
	for(int i=1;i<=n;i++)
	{
		if(!dfn[i])
		    tarjan(i,-1);
	}
}
void suodian()
{
	int u,v;
	memset(in,0,sizeof(in));
	for(int i=0;i<ans;i+=2)
	{
		u=ebcno[edge[i].beg];
		v=ebcno[edge[i].end];
		if(u!=v)
		{
			newmap[u].push_back(v);
			newmap[v].push_back(u);
			in[u]++;
			in[v]++;
		}
	}
}
void solve()
{
	int i,j,sum=0;;
	for(i=1;i<=ebccnt;i++)
	{
		if(in[i]==1)
		    sum++;
	}
	printf("%d\n",(sum+1)/2);
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		init();
		getmap();
		find();
		suodian();
		solve();
	}
	return 0;
}