The Stable Marriage Problem

Posted 2021-02-09 bxd123

和上一题基本一样的 

#include <iostream>
#include <cstring>
#include <map>
#define MAXN 30
using namespace std;
int mm[MAXN][MAXN], gg[MAXN][MAXN], p[MAXN];
int gg_match[MAXN], mm_match[MAXN];
char idx[2*MAXN];
map<char, int> mp;
int main() {
    int t, n;
    cin >> t;
    while (t--) {
        cin >> n;
        char ch;
        int cnt = 0;
        mp.clear();
        for (int i = 0; i < 2 * n; i++) cin >> ch;
        for (int i = 1; i <= n; i++) {
            cin >> idx[i] >> ch;
            mp[idx[i]] = i;
            for (int j = 1; j <= n; j++) {
                cin >> ch;
                if (mp[ch] == 0) {
                    mp[ch] = ++cnt;
                    idx[cnt+n] = ch;
                }
                gg[i][j] = mp[ch];
            }
        }
        for (int i = 1; i <= n; i++) {
            char gir;
            cin >> gir >> ch;
            int a = mp[gir];
            for (int j = 1; j <= n; j++) {
                cin >> ch;
                int b = mp[ch];
                mm[a][b] = j;
            }
        }
        bool flag = true;
        for (int i = 1; i <= n; i++) p[i] = 1;
        memset(mm_match, -1, sizeof(mm_match));
        memset(gg_match, -1, sizeof(gg_match));
        while (flag) {
            flag = false;
            for (int i = 1; i <= n; i++) {
                if (gg_match[i] == -1 && p[i] <= n) {
                    int a = gg[i][p[i]++];
                    if (mm_match[a] == -1) mm_match[a] = i, gg_match[i] = a;
                    else if (mm[a][i] < mm[a][mm_match[a]]) {
                        gg_match[i] = a;
                        gg_match[mm_match[a]] = -1;
                        mm_match[a] = i;
                    }
                    flag = true;
                }
            }
        }
        for (int i = 0; i < 26; i++) {
            for (int j = 1; j <= n; j++) if (idx[j] == ‘a‘ + i) {
                cout << idx[j] << ‘ ‘ << idx[gg_match[j]+n] << endl;
                break;
            }
        }
        if (t) cout << endl;
    }
    return 0;
}