1276B

dfs

本质上这道题是求割点所在子树大小，但是由于只有两个点，所以直接$dfs$求即可

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, a, b, ca, cb;
int vis[maxn];
vector<int> G[maxn];
void dfs_1(int u) {
    if(vis[u]) {
        return;
    }
    vis[u] = 1;
    ++ca;
    for(auto v : G[u]) {
        dfs_1(v);
    }
} 
void dfs_2(int u) {
    if(vis[u]) {
        return;
    }
    vis[u] = 1;
    ++cb;
    for(auto v : G[u]) {
        dfs_2(v);
    }
}
int main() {
    int T; scanf("%d", &T);
    while(T--) {
        ca = cb = 0;
        scanf("%d%d%d%d", &n, &m, &a, &b);
        for(int i = 1; i <= n; ++i) {
            G[i].clear();
        }
        for(int i = 1; i <= m; ++i) {
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(int i = 1; i <= n; ++i) {
            vis[i] = 0;
        }
        vis[b] = 1; 
        dfs_1(a);
        for(int i = 1; i <= n; ++i) {
            vis[i] = 0;
        }
        vis[a] = 1;
        dfs_2(b);
        printf("%lld\n", 1LL * (n - ca - 1) * (n - cb - 1));
    }
    return 0;
}