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;
}
2.矩阵变换
不知道是怎么发现的一个性质
对于一行，它更喜欢这一行靠前的数
对于一个数，它更喜欢它的位置靠后的行
然后就转化为一个稳定婚姻问题了
关于上面的那条性质，这是一个例子：
0 1 0 2
1 0 2 0
行\(1\)喜欢的数的顺序：\(1\ 2\)
行\(2\)喜欢的数的顺序：\(1\ 2\)
数\(1\)对于行的喜欢顺序：\(1\ 2\)
数\(2\)对于行的喜欢顺序：\(1\ 2\)
如果行\(1\)和数\(2\)配对的话，就会出现行\(1\)更喜欢数\(1\)且数\(1\)更喜欢行\(1\)的情况，不合法，稳定婚姻算法可以避免这种情况的出现
代码：
#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 200
#define M 200
int n, m;
int matchx[N+5], matchy[N+5], a[N+5][2*N+5];
int pref[N+5][N+5], order[N+5][N+5], nxt[N+5], head[N+5];
queue<int> q;
void engage(int man, int woman) {
int m = matchy[woman];
if(m) {
matchx[m] = 0;
q.push(m);
}
matchx[man] = woman;
matchy[woman] = man;
}
int main() {
int T;
scanf("%d", &T);
while(T--) {
scanf("%d%d", &n, &m);
while(!q.empty()) q.pop();
for(int i = 1; i <= n; ++i) {
nxt[i] = matchx[i] = matchy[i] = head[i] = 0;
int tot = 0;
for(int j = 1; j <= m; ++j) {
scanf("%d", &a[i][j]);
if(!a[i][j]) continue;
pref[i][++tot] = a[i][j];
}
q.push(i);
}
for(int j = m; j >= 1; --j)
for(int i = 1; i <= n; ++i)
if(a[i][j]) order[a[i][j]][i] = ++head[a[i][j]];
while(!q.empty()) {
int u = q.front(); q.pop();
int v = pref[u][++nxt[u]];
if(!matchy[v]) engage(u, v);
else if(order[v][u] < order[v][matchy[v]]) engage(u, v);
else q.push(u);
}
for(int i = 1; i <= n; ++i) printf("%d ", matchx[i]);
printf("\n");
}
return 0;
}
