poj 3177 3352 边双连通分量

题目链接：http://poj.org/problem?id=3177

题意：至少加多少边才能构成边双连通图。

思路：求边双连通分量，对于每一个边双连通分量都看做一个点，也就是进行缩点，然后就构造出一个树，然后在这个树上加多少个树边，能够变成一个双连通图。 加的数量就是（ 这棵树总度数为1的结点数 + 1 ）/ 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
const int M_node = 5009,M_edge = 20009;
struct edge
{
    int to,next;
    bool cut;
}edge[M_edge];
stack<int> st;
int head[M_node],tot;
int low[M_node],dfn[M_node],belong[M_node];
int degree[M_node];
int dfs_clock,block,bridge;
bool instack[M_node];
int n,m;
void init()
{
    while(!st.empty()) st.pop();
    memset(head,-1,sizeof(head));
    memset(degree,0,sizeof(degree));
    tot = 0;
    memset(instack,false,sizeof(instack));
    memset(belong,-1,sizeof(belong));
    dfs_clock = 0;
    block = bridge = 0;
}
void add_edge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].cut = false;
    head[u] = tot++;
}
void tarjan(int u,int fa)
{
    low[u] = dfn[u] = ++dfs_clock;
    st.push(u);
    instack[u] = true;
    for(int i = head[u]; i != -1 ;i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == fa) continue;
        if(!dfn[v])
        {
            tarjan(v,u);
            low[u] = min(low[u],low[v]);
            if(low[v] > dfn[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
        }
        else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];
    }
    if(low[u] == dfn[u])
    {
        block++;
        for(;;)
        {
            int v = st.top();
            st.pop();
            instack[v] = false;
            belong[v] = block;
            if(v == u) break;
        }
    }
}
void solve()
{
    int ans = 0;
    tarjan(1,-1);
    for(int i = 1;i <= n;i++)
    {
        for(int j = head[i];j != -1;j = edge[j].next)
        {
            if(edge[j].cut) degree[belong[i]]++;
        }
    }
    for(int i = 1;i <= block;i++)
    {
        if(degree[i] == 1) ans++;
    }
    printf("%d\n",(ans+1)/2);
}
int main()
{
    while(scanf("%d %d",&n,&m) == 2)
    {
        init();
        for(int i = 0;i < m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            add_edge(a,b);
            add_edge(b,a);
        }
        solve();
    }
    return 0;
}