【电力——tarjan割点，求连通块】

题目

分析

这是割点的板子 

代码

#include <bits/stdc++.h>
using namespace std;

const int N = 1e4+10;
const int M = 3e4+10;

int h[N], e[M], ne[M], idx;
int dfn[N], low[N], tot;
int root, ans;

void add(int a, int b)  // 添加一条边a->b
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++tot;
    
    int cnt = 0;
    for(int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
            if(low[j] >= dfn[u])
                cnt++;
        }
        else low[u] = min(low[u], dfn[j]);
    }
    
    if(u != root) cnt++;
    ans = max(ans, cnt);
}
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m), n || m)
    {
        idx = tot = 0;
        memset(h, -1, sizeof h);
        memset(dfn, 0, sizeof dfn);
        for(int i = 1; i <= m; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            add(a, b), add(b, a);
        }
        
        ans = 0; //删掉结点的最大贡献
        int cnt = 0; //连通块数目
        for(root = 0; root < n; root++)
            if(!dfn[root])
                tarjan(root), cnt++;
                
        printf("%d\n", cnt + ans - 1);
    }
}