Identifying Infected

\quad对于每次查询(a,b)，都去循环遍历图中每个点f，假设去掉这个点后a是否还能到b，如果去掉f后a不能到b，则f点一定被感染。这样的话时间复杂度为O(Q∗N∗(N+M))O(Q*N*(N+M))O(Q∗N∗(N+M))，可以过掉30%的数据。

#include <iostream>
#include <queue>
#include <vector>
using namespace std;const int N = 1010;
vector<int> E[N];
int vis[N];
int n, m;
// 看图中是否存在从a到b且不经过f的路径
bool haspath(int a, int b, int f)
{for(int i = 1; i <= n; i ++ ) vis[i] = 0;queue<int> q;q.push(a);vis[a] = 1;while(!q.empty()){int x = q.front();q.pop();if(x == b) return true;for(int y: E[x]){if(vis[y] == 1 || y == f) continue;q.push(y);vis[y] = 1;}}return false;
}int main()
{cin >> n >> m;for(int i = 0; i < m; i ++ ){int u, v; cin >> u >> v;E[u].push_back(v);E[v].push_back(u);}int Q; cin >> Q;while(Q -- ){int u, v; cin >> u >> v;int res = 2;// 依次去掉除了起点和终点的其他点后看看图中是否还存在一条从u到v的路，如果不存在的话则这个点必经过，需要在答案中+1for(int i = 1; i <= n; i ++ )if(i != u && i != v && !haspath(u, v, i)) res ++ ;cout << res << endl;}return 0;
}

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