UESTC 899 方老师和农场 --双连通分量的构造

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <time.h>
#include <queue>
#include <cctype>
#include <utility>
#include <numeric>
#include <cstdlib>
#include <functional>
#include <iomanip>
#include <sstream>
#define Mod 1000000007
#define SMod 10007
#define INT 2147483647
#define pi acos(-1.0)
#define eps 1e-4
#define lll __int64
#define ll long long
using namespace std;
#define N 10007

vector<int> G[N];
struct Bridge
{
    int u,v;
}bg[2*N];

int vis[N],low[N],dfn[N],Time;
int fa[N],deg[N];
int n,m,cnt;

int findset(int x)
{
    if(x != fa[x])
        fa[x] = findset(fa[x]);
    return fa[x];
}

void Tarjan(int u,int father)
{
    low[u] = dfn[u] = ++Time;
    vis[u] = 1;
    for(int i=0;i<G[u].size();i++)
    {
        int v = G[u][i];
        if(v == father)
            continue;
        if(!vis[v])
        {
            Tarjan(v,u);
            low[u] = min(low[u],low[v]);
            if(low[v] > dfn[u])        //u->v为桥
                bg[cnt].u = u,bg[cnt++].v = v;
            else   //否则，u,v同属一个连通分量，合并
            {
                int fx = findset(u);
                int fy = findset(v);
                if(fx != fy)
                    fa[fx] = fy;
            }
        }
        else
            low[u] = min(low[u],dfn[v]);
    }
}

int main()
{
    int i,j,u,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<=n;i++)
            G[i].clear();
        cnt = Time = 0;
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(i=1;i<=n;i++)
            fa[i] = i;
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(vis,0,sizeof(vis));
        Tarjan(1,-1);
        //统计桥端度数
        memset(deg,0,sizeof(deg));
        for(i=0;i<cnt;i++)
        {
            int fx = findset(bg[i].u);  //fx,fy为缩点后的代表点
            int fy = findset(bg[i].v);
            deg[fx]++;
            deg[fy]++;
        }
        int leaf = 0;
        for(i=1;i<=n;i++)
            if(deg[i] == 1)
                leaf++;
        printf("%d\n",(leaf+1)/2);
    }
    return 0;
}