Va 1175-Ladies 'choice (stable marital problem)

Va 1175-Ladies 'choice

Question Link

Given n men and N women, each person has a ranking of the opposite sex. From left to right, they like most to dislike most. Then, a stable match is required, so that N men and women do not have men who are more emotional than other women and women who are more emotional than other men

Idea: The algorithm process for stable marriage is as follows:
A man keeps proposing proposal, from the favorite to the least. Every time a woman chooses a favorite pairing among the propose, the current pairing is discarded, this process can be used to store the proposal man in a queue, so that the matching is completed until the queue is empty.

Code:

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int N = 1005;int t, n;struct People {int order[N], now, match;} man[N], woman[N];queue<int> Q;bool gao(int u, int v) {if (woman[v].match == -1) {woman[v].match = u;return true;} else if (woman[v].order[woman[v].match] > woman[v].order[u]) {man[woman[v].match].now++;Q.push(woman[v].match);woman[v].match = u;return true;}return false;}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);for (int i = 0; i < n; i++) {man[i].now = 0;for (int j = 0; j < n; j++) {scanf("%d", &man[i].order[j]);man[i].order[j]--;}Q.push(i);}for (int i = 0; i < n; i++) {woman[i].match = -1;int tmp;for (int j = 0; j < n; j++) {scanf("%d", &tmp);tmp--;woman[i].order[tmp] = j;}}while (!Q.empty()) {int u = Q.front();Q.pop();int v = man[u].order[man[u].now];if (!gao(u, v)) {man[u].now++;Q.push(u);}}for (int i = 0; i < n; i++)man[woman[i].match].match = i;for (int i = 0; i < n; i++)printf("%d\n", man[i].match + 1);if (t) printf("\n");}return 0;}

Va 1175-Ladies 'choice (stable marital problem)