稳定婚姻问题学习笔记

Gale-Shapley Algorithm

此算法的流程如下：
首先搞一个队列，存储未匹配的男士编号。每次取出一个未匹配的男士的编号，让他向其未求过婚的且最喜欢的女士求婚，如果对应女士没有匹配或者已经匹配的没有这位优，那么将这位与对应的女士相匹配，并且将原来已匹配的男士扔到队列里，一直重复上述步骤直到队列为空
可以证明，上述流程结束后，每位男士必定有配偶，且婚姻是稳定的

例题

1.UVA1175 Ladies' Choice
最简单的板子，直接上代码，抄的大刘的

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define DEF 0x8f8f8f8f
#define DDEF 0x8f8f8f8f8f8f8f8fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline

#define N 1000
int pref[N+5][N+5], order[N+5][N+5], nxt[N+5];
int futureHusband[N+5], futureWife[N+5];
queue<int> q;

void engage(int man, int woman) {
  int m = futureHusband[woman];
  if(m) {
    futureWife[m] = 0;
    q.push(m);
  }
  futureWife[man] = woman;
  futureHusband[woman] = man;
}

int main() {
  int T;
  scanf("%d", &T);
  while(T--) {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
      for(int j = 1; j <= n; ++j) scanf("%d", &pref[i][j]);
      nxt[i] = 1;
      futureWife[i] = 0;
      q.push(i);
    }
    for(int i = 1; i <= n; ++i) {
      for(int j = 1; j <= n; ++j) {
        int x;
        scanf("%d", &x);
        order[i][x] = j;
      }
      futureHusband[i] = 0;
    }
    while(!q.empty()) {
      int man = q.front(); q.pop();
      int woman = pref[man][nxt[man]++];
      if(!futureHusband[woman]) engage(man, woman);
      else if(order[woman][man] < order[woman][futureHusband[woman]]) engage(man, woman);
      else q.push(man);
    }
    while(!q.empty()) q.pop();
    for(int i = 1; i <= n; ++i) printf("%d\n", futureWife[i]);
    if(T) printf("\n");
  }
  return 0;
}