链接：https://vjudge.net/problem/POJ-3352#author=0
题意：
给一个无向连通图，至少添加几条边使得去掉图中任意一条边不改变图的连通性（即使得它变为边双连通图）。
思路：
将图中的边双联通分量全部缩成一个点，得到度为1的点的数目。
若要使缩点后的图都边双联通，增加(leaf+1)/2条边即可。leaf就是度为1的点。
代码：
#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e3+10;
vector<int> G[MAXN];
stack<int> St;
int Dfn[MAXN], Low[MAXN];
int Dis[MAXN], Fa[MAXN];
int times, res, cnt;
int n, m;
void Init()
{
for (int i = 1;i <= n;i++)
G[i].clear();
memset(Dfn, 0, sizeof(Dfn));
memset(Low, 0, sizeof(Low));
memset(Dis, 0, sizeof(Dis));
memset(Fa, 0, sizeof(Fa));
times = res = cnt = 0;
}
void Tarjan(int u, int v)
{
Dfn[v] = Low[v] = ++times;
St.push(v);
for (int i = 0;i < G[v].size();i++)
{
int node = G[v][i];
if (node == u)
continue;
if (Dfn[node] == 0)
Tarjan(v, node);
Low[v] = min(Low[v], Low[node]);
}
if (Dfn[v] == Low[v])
{
cnt++;
int node;
do
{
node = St.top();
Fa[node] = cnt;
St.pop();
}
while (node != v);
}
}
int main()
{
string s;
int t;
while (cin >> n >> m)
{
int l, r;
// cin >> n >> m;
Init();
for (int i = 1;i <= m;i++)
{
cin >> l >> r;
G[l].push_back(r);
G[r].push_back(l);
}
Tarjan(0, 1);
// copy(Fa+1, Fa+1+n, ostream_iterator<int> (cout, " "));
// copy(Dis+1, Dis+1+n, ostream_iterator<int> (cout, " "));
// cout << endl;
for (int i = 1;i <= n;i++)
{
for (int j = 0;j < G[i].size();j++)
{
int node = G[i][j];
if (node == i)
continue;
if (Fa[i] != Fa[node])
Dis[Fa[node]]++;
}
}
int leaf = 0;
for (int i = 1;i <= cnt;i++)
{
if (Dis[i] == 1)
leaf++;
}
// cout << "Output for Sample Input " << t << endl;
cout << (leaf+1)/2 << endl;
}
return 0;
}