题意：
A和B告白，必须通过A的同性朋友和B的同性朋友传达，要求输出所以可能的两个朋友的情况
思路：
这题有些复杂，我一开始想复杂了，上来就用dfs+回溯做，结果超时，其实只要遍历所有A的朋友和B的朋友，看这两者是否互为朋友即可。不过这题还有很多地方要注意，我还是看别人的代码写的。
https://www.liuchuo.net/archives/4210
#include<iostream>
#include<algorithm>
#include<vector>
#include<functional>
#include<string>
#include<cmath>
#include<queue>
#include<map>
#include<unordered_map>
#include<set>
using namespace std;
const int maxn = 10050;
const int inf = 10000;
unordered_map<int, bool> arr;
vector<int> v[maxn];
struct node {int a, b;
};bool cmp(node x, node y) {if (x.a != y.a) {return x.a < y.a;} else {return x.b < y.b;}
}int main() {int n, m, k;scanf("%d%d", &n, &m);for (int i = 0; i < m; i++) {string a, b;cin >> a >> b;if (a.length() == b.length()) {v[abs(stoi(a))].push_back(abs(stoi(b)));v[abs(stoi(b))].push_back(abs(stoi(a)));}arr[abs(stoi(a))* inf + abs(stoi(b))] = arr[abs(stoi(b))* inf+ abs(stoi(a))] = true;}scanf("%d", &k);for (int i = 0; i < k; i++) {int c, d;cin >> c >> d;vector<node> ans;for (int j = 0; j < v[abs(c)].size(); j++) {for (int l = 0; l < v[abs(d)].size(); l++) {int s = v[abs(c)][j];int e = v[abs(d)][l];if (s == abs(d) || e == abs(c))continue;if (arr[s*inf + e] == true) {ans.push_back(node{ s,e });}}}sort(ans.begin(), ans.end(), cmp);printf("%d\n", ans.size());for (int j = 0; j < ans.size(); j++) {printf("%04d %04d\n", ans[j].a, ans[j].b);}}return 0;
}