Codeforces Round #606 (Div. 2) - E. Two Fairs(割点+dfs)

 

 

 

#include <iostream>
#include <algorithm>
#include <cstdio>

using namespace std;

const int N = 500010;

typedef long long ll;

struct node {
    int to, nex;
};

node edge[2 * N];
int t, n, m, a, b, cnt, head[N];
int dfn[N], timing, pcut[N];
int cnta, cntb, vis[N];

void dfs(int u, int mk, int a, int b)
{
    vis[u] = 1;
    if (0 == mk) cnta++;
    if (1 == mk) cntb++;
    if (0 == mk && u == b) return;
    if (1 == mk && u == a) return;
    for (int i = head[u]; 0 != i; i = edge[i].nex) {
        int v = edge[i].to;
        if (!vis[v]) dfs(v, mk, a, b);
    }
    return;
}

void add_edge(int u, int v)
{
    edge[++cnt].to = v;
    edge[cnt].nex = head[u];
    head[u] = cnt;
}

int tarjan(int u, int fa)
{
    int child = 0, lowu;
    lowu = dfn[u] = ++timing;
    for (int i = head[u]; 0 != i; i = edge[i].nex) {
        int v = edge[i].to;
        if (!dfn[v]) {
            child++;
            int lowv = tarjan(v, u);
            if (lowv >= dfn[u] && u != fa) pcut[u] = 1;
            lowu = min(lowu, lowv);
        }
        else if (v != fa) {
            lowu = min(lowu, dfn[v]);
        }
    }
    if (u == fa && child > 1) pcut[u] = 1;
    return lowu;
}

void init()
{
    cnt = timing = cnta = cntb = 0;
    for (int i = 1; i <= n; i++)
        pcut[i] = dfn[i] = head[i] = 0;
}

int main()
{
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d%d", &n, &m, &a, &b);
        init();
        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            add_edge(u, v), add_edge(v, u);
        }
        for (int i = 1; i <= n; i++) {
            if (0 == dfn[i]) tarjan(i, i);
        }
        if (0 == pcut[a] || 0 == pcut[b]) {
            printf("0\n");
            continue;
        }
        for (int i = 1; i <= n; i++) vis[i] = 0;
        dfs(a, 0, a, b);
        for (int i = 1; i <= n; i++) vis[i] = 0;
        dfs(b, 1, a, b);
        ll x = ll(n - cnta), y = ll(n - cntb);
        printf("%lld\n", x * y);
    }
    return 0;
}