uva 10972 RevolC FaeLoN (双连通分量)

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=54643#problem/H

题意：一个无向图要添加多少条边才能使其变为边双连通分量

思路：可以将缩点后的图画出，用类似欧拉回路的思想求解，如果某点度数为1（叶子点）,那么ans + 1, 如果是孤立点，那么ans + 2, 最后的结果便是 (ans + 1) / 2 （加一为了向上取整），如果一开始本图就是强联通分量，那就输出0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

const int N = 2020;

int n, tot;
int dfn
, low
, ins
, belong
, de
, dcnt, bcnt;
vector <int> e
;
stack <int> sta;
typedef pair <int, int> P;
vector <P> bridge;

void init()
{
dcnt = bcnt = 0;
while(!sta.empty()) sta.pop();
bridge.clear();
for(int i = 1; i <= n; i++)
{
dfn[i] = 0;
e[i].clear();
ins[i] = de[i] = 0;
}
}

void dfs(int u , int fa)
{
sta.push(u);
ins[u] = true;
dfn[u] = low[u] = ++dcnt;
for(int i = 0; i < e[u].size(); i++)
{
int v = e[u][i];
if(v == fa) continue;
if(!dfn[v]) //树边
{
dfs(v, u);
low[u] = min(low[u] , low[v]);
if(low[v] > dfn[u]) //桥
{
bridge.push_back(make_pair(u, v));
++bcnt;
while(true)
{
int x = sta.top();
sta.pop();
ins[x] = 0;
belong[x] = bcnt;
if(x == v) break;
}
}
}
else
if(ins[v]) //后向边
low[u] = min(low[u] , dfn[v]);
}
}

void solve()
{
for(int i = 1; i <= n; i++)
if(!dfn[i])
{
dfs(i, -1);
bcnt++;
while(!sta.empty())
{
int x = sta.top();
sta.pop();
ins[x] = 0;
belong[x] = bcnt;
if(x == i) break;
}
}
if(bcnt == 1)
{
cout << 0 << endl;
return ;
}
for(int i = 0; i < bridge.size(); i++) //取出所有的桥
{
int u = bridge[i].first;
int v = bridge[i].second;
de[belong[u]]++;
de[belong[v]]++;
}
int res = 0;
for(int i = 1; i <= bcnt; i++)
if(de[i] == 0)
res += 2;
else
if(de[i] == 1)
res++;
cout << (res + 1) / 2 << endl;
}

int main()
{
while(cin >> n >> tot)
{
init();
while(tot--)
{
int u, v;
cin >> u >> v;
e[u].push_back(v);
e[v].push_back(u);
}
solve();
}
return 0;
}