Codeforces - Two Fairs

There are n cities in Berland and some pairs of them are connected by two-way roads. It is guaranteed that you can pass from any city to any other, moving along the roads. Cities are numerated from 1 to n.
Two fairs are currently taking place in Berland — they are held in two different cities a and b (1≤a,b≤n; a≠b).
Find the number of pairs of cities x and y (x≠a,x≠b,y≠a,y≠b) such that if you go from x to y you will have to go through both fairs (the order of visits doesn’t matter). Formally, you need to find the number of pairs of cities x,y such that any path from x to y goes through a and b (in any order).
Print the required number of pairs. The order of two cities in a pair does not matter, that is, the pairs (x,y) and (y,x) must be taken into account only once.
Input The first line of the input contains an integer t (1≤t≤4⋅104) — the number of test cases in the input. Next, t test cases are specified.
The first line of each test case contains four integers n, m, a and b (4≤n≤2⋅105, n−1≤m≤5⋅105, 1≤a,b≤n, a≠b) — numbers of cities and roads in Berland and numbers of two cities where fairs are held, respectively.
The following m lines contain descriptions of roads between cities. Each of road description contains a pair of integers ui,vi (1≤ui,vi≤n, ui≠vi) — numbers of cities connected by the road.
Each road is bi-directional and connects two different cities. It is guaranteed that from any city you can pass to any other by roads. There can be more than one road between a pair of cities.
The sum of the values of n for all sets of input data in the test does not exceed 2⋅105. The sum of the values of m for all sets of input data in the test does not exceed 5⋅105.
Output Print t integers — the answers to the given test cases in the order they are written in the input.
Example inputCopy 3 7 7 3 5 1 2 2 3 3 4 4 5 5 6 6 7 7 5 4 5 2 3 1 2 2 3 3 4 4 1 4 2 4 3 2 1 1 2 2 3 4 1 outputCopy 4 0 1
제목의 대의: 몇 가지 옳은 점을 찾아내고 한 점을 만족시키면 다른 점까지 반드시 a, b 두 점을 지나갈 것이다.조건에 맞는 점 쌍 수를 출력합니다.
그래서 우리는 반드시 a를 통과해야만 b에 도착할 수 있는 개수와 반드시 b를 통과해야만 a에 도착할 수 있는 개수를 찾아야 한다.그리고 곱하면 된다.
어떻게 구하지?반드시 a를 통과해야만 b에 도착할 수 있다. 실제로 우리는 b점에 대해 bfs를 뛴 다음에 a점을 통과할 수 없도록 제한한다. 그러면 도착하지 못하는 점이 얼마나 많은지 보면 된다.후자는 같다.
AC 코드:

#include
#define int long long
using namespace std;
const int N=2e5+10;
int T,n,m,a,b,res,vis[N],cnt;
vector<int> v[N];
inline void add(int a,int b){v[a].push_back(b); v[b].push_back(a);}
inline int bfs(int x,int y){
    queue<int> q;   q.push(x);  cnt=2;  memset(vis,0,8*(n+1)); vis[x]=vis[y]=1;
    while(q.size()){
        int u=q.front();    q.pop();
        for(int i=0,to;i<v[u].size();i++){
            to=v[u][i];
            if(!vis[to])    vis[to]=1,q.push(to),cnt++;
        }
    }
    return n-cnt;
}
signed main(){
    ios::sync_with_stdio(false);  cin.tie(nullptr); cout.tie(nullptr);
    cin>>T;
    while(T--){
        cin>>n>>m>>a>>b;
        for(int i=1,x,y;i<=m;i++)   cin>>x>>y,add(x,y);
        res=bfs(a,b); 	res*=bfs(b,a);
        cout<<res<<'
';
        for(int i=1;i<=n;i++)   v[i].clear();
    }
    return 0;
}