poj 3177 Redundant Paths(边双连通分量+缩点)

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https://cn.vjudge.net/problem/POJ-3177

题意

问最少能补成几条边，使这个图成为边双连通

思路

https://www.cnblogs.com/chenchengxun/p/4718736.html

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 1000000
vector<int>mp[maxn];
int degree[maxn]; //记录度
int block[maxn];
int Stack[maxn];
int dfn[maxn];
int low[maxn];
int Father[maxn];
int n,m,cnt,top,tt;

void init()
{
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(degree,0,sizeof(degree));
    memset(Stack,0,sizeof(Stack));
    memset(Father,0,sizeof(Father));
    top = tt = cnt = 0;
    for(int i = 1;i<=n;i++) mp[i].clear();
}

void Tarjan(int u,int fa)
{
    dfn[u] = low[u] = ++tt;
    Father[u] = fa;
    Stack[top++] =  u;
    int k = 0;
    for(int i = 0;i< mp[u].size();i++)
    {
        int v = mp[u][i];
        if(v == fa&&!k)
        {
            k++;
            continue;
        }
        if(low[v] == 0)
        {
            Tarjan(v,u);
            low[u] = min(low[u],low[v]);
        }
        else low[u] = min(low[u],low[v]);
    }
    int v;
    if(dfn[u] == low[u])//找到强连通图
    {
        do
        {
            v = Stack[--top];
            block[v] = cnt;
        }
        while(u != v); //染色
        cnt++;
    }
}

void solve()
{
    memset(degree,0,sizeof(degree));
    for(int i=1;i<=n;i++)
    {
        if(low[i] == 0)
        {
            Tarjan(i,i);
        }
    }
   for(int i  = 1;i<=n;i++)
   {
       int v= Father[i];
       if(block[i] != block[v])
       {
           degree[block[i]]++;
           degree[block[v]]++;
       }
   }
   int ans=0;
   for(int i=0;i<cnt;i++)
   {
       if(degree[i] == 1)
       {
           ans++;
       }
   }
   printf("%d\n",(ans+1)/2);
}

int main()
{
    while(scanf("%d %d",&n ,&m) != EOF)
    {
        init();
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            mp[x].push_back(y);
            mp[y].push_back(x);
        }
        solve();
    }
}