BZOJ 1123 tarjan

题目链接

题意：一张无向图，把第$i$个点关联的所有边去掉，求无向图中有多少个点对不连通。

题解：

如果割的不是割点，那么总答案是$2\times (n-1)$.

如果是割点，要分别考虑每个子树的贡献.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int N = 1e5 + , M = 5e5 + ;
int head[N], ver[M * ], Next[M * ];
int dfn[N], low[N], Size[N];
ll ans[N];
bool cut[N];
int n, m, tot, num; ///时间戳号

void add(int u, int v)
{
    ver[++tot] = v, Next[tot] = head[u], head[u] = tot;
}
void tarjan(int u)
{
    dfn[u] = low[u] = ++num; Size[u] = ;
    int flag = , sum = ;
    for(int i = head[u]; i; i = Next[i])
    {
        int v = ver[i];
        if(!dfn[v])
        {
            tarjan(v);
            Size[u] += Size[v];
            low[u] = min(low[u], low[v]);
            if(low[v] >= dfn[u])
            {
                flag++;
                ans[u] += (ll)Size[v] * (n - Size[v]);
                sum += Size[v];
                if(u !=  || flag > ) cut[u] = true;
            }
        }
        else low[u] = min(low[u], dfn[v]);
    }
    if(cut[u]) ans[u] += (ll)(n - sum - ) * (sum + ) + (n - );
    else ans[u] =  * (n - );
}
int main()
{
    scanf("%d %d", &n, &m); tot = ;
    for(int i = ; i <= m; i++)
    {
        int u, v; scanf("%d %d", &u, &v);
        if(u == v) continue;
        add(u, v), add(v, u);
    }
    tarjan();
    for(int i = ; i <= n; i++) printf("%lld\n", ans[i]);
    return ;
}

Code