uva 10972(边双连通分量）

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=33804、

思路：和poj的一道题有点像，不过这道题图可能不连通，因此首先求边双连通分量，然后算每个连通分量的度数，显然叶子节点的度数为1，孤立点的度数为0，然后就是统计度数了，对于孤立点ans+=2,对于叶子节点，ans++。于是最后的答案就是(ans+1)/2了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <stack>
 #include <vector>
 using namespace std;
 #define MAXN 1111

 int n, m, cnt, _count;
 stack <int >S;
 vector <vector<int > >g;

 int low[MAXN], dfn[MAXN], color[MAXN];
 int degree[MAXN];
 bool mark[MAXN];
 void Tarjan(int u, int father)
 {
     low[u] = dfn[u] = ++ cnt;
     S.push(u);
     mark[u] = true;
     for(int i = ; i < g[u].size(); i ++ ){
         int v = g[u][i];
         if(v == father)continue;
         if(dfn[v] == ) {
             Tarjan(v, u);
             low[u] = min(low[u], low[v]);
         } else if(mark[v]) {
             low[u] = min(low[u], dfn[v]);
         }
     }
     if(low[u] == dfn[u]){
         int x;
         _count++;
         do {
             x = S.top();
             S.pop();
             mark[x] = false;
             color[x] = _count;
         }while(x != u);
     }
 }

 int main()
 {
     int u, v, ans;
     while(~scanf("%d %d", &n, &m)){
         g.clear();
         g.resize(n+);
         while(m --){
             scanf("%d %d",&u, &v);
             g[u].push_back(v);
             g[v].push_back(u);
         }
         memset(dfn, , sizeof(dfn));
         memset(mark, false, sizeof(mark));
         cnt = _count = ;
         for(int i = ; i <= n; i ++){
             if(dfn[i] == )Tarjan(i, -);
         }
         if(_count == ){
             puts("");
             continue;
         }
         memset(degree, , sizeof(degree));
         for(int i = ; i <= n; i++){
             for(int j = ; j < g[i].size(); j++){
                 if(color[i] != color[g[i][j]])degree[color[g[i][j]]] ++;
             }
         }
         ans = ;
         for(int i = ; i <= _count; i++){
             if(degree[i] == )ans += ; // 孤立点
             else if(degree[i] == )ans ++; // 叶子节点
         }
         printf("%d\n", (ans + )/ );
     }
     return ;
 }