题目链接
稳定婚姻问题：
男方向依次向女方求爱，如果女方没有配偶，则配对，如果女方有配偶，比较配偶和当前求爱的男方，选择好的进行配对，反复如此，直到所有的男方都配对完成。
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <math.h>
#include <bitset>
#include <ctype.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 225 + 5;
const int mod = 1e9 + 7;
int t, kase = 0;
int n;
vector<int> G[N];
int match[N];
void init()
{for(int i = 0; i < N; i++) G[i].clear();
}void solve()
{memset(match, 0, sizeof(match));queue<int> Q;for(int i = 1; i <= n; i++)Q.push(i);while(!Q.empty()){int u = Q.front(); Q.pop();if(match[u]) continue;int m = G[u].size();for(int i = 0; i < m; i++){int v = G[u][i];if(!match[v]){match[v] = u;match[u] = v;break;}else{int pm = G[v].size();for(int j = 0; j < m; j++){int pu = G[v][j];if(pu == match[v])break;if(pu == u){match[u] = v;int tu = match[v];match[tu] = 0;Q.push(tu);match[v] = u;break;}}if(match[u]) break;}}}
}int main()
{scanf("%d", &t);while(t--){init();scanf("%d", &n);for(int i = 1; i <= 2*n; i++){for(int j = 1; j <= n; j++){int tmp;scanf("%d", &tmp);G[i].push_back(tmp);}}solve();printf("Case %d: ", ++kase);for(int i = 1; i <= n; i++){if(i != 1) printf(" ");printf("(%d %d)", i, match[i]);}printf("\n");}return 0;
}