【稳定婚姻问题】poj3487 The Stable Marriage Problem

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具体参见上篇博客。

都是类似的问题，不过我发现这类问题好像输入格式的控制会稍微难一点。

Gale-shapley的实现会相对简单很多。其中求loc部分其实也可以用map优化一下，否则我的实现是可能退化到O(n^3)的。

当然这问题规模实在是太小了，所以也就不需要优化。。。

#include <cstdio>#include <cstring>#include <vector>#include <utility>#include <set>#include <map>#include <algorithm>using namespace std;const int MAX = 30;int ggid[MAX], mmid[MAX];vector<int> gg[MAX], mm[MAX];char gglist[MAX], mmlist[MAX];bool vis[MAX];bool rej[MAX][MAX];int match[MAX];int n;int loc(int a, int b) {for (int i = 0; i < n; ++i) {if (mm[a][i] == b) return i;}}void solve() {memset(vis, false, sizeof(vis));memset(match, -1, sizeof(match));memset(rej, false, sizeof(rej));bool flag = true;while (flag) {flag = false;for (int i = 1; i <= n; ++i) {if (!vis[i]) {for (int j = 0; j < n; ++j) {int &e = gg[i][j];if (match[e] == -1) {vis[i] = true;match[e] = i;//printf("1: %c match %c\n", gglist[i], mmlist[e]);break;} else if (!rej[e][i] && loc(e, match[e]) > loc(e, i)) {vis[match[e]] = false;vis[i] = true;match[e] = i;//printf("2: %c match %c\n", gglist[i], mmlist[e]);break;} else {rej[e][i] = true;}}flag = true;}}}set<pair<char, char> > ans;for (int i = 1; i <= n; ++i) {ans.insert(make_pair(gglist[match[i]], mmlist[i]));}for (set<pair<char, char> >::iterator it = ans.begin(); it != ans.end(); ++it) {printf("%c %c\n", it->first, it->second);}}int main() {int T;scanf(" %d", &T);while (T--) {scanf(" %d", &n);//initfor (int i = 0; i < MAX; ++i) {gg[i].clear();mm[i].clear();}/*gglist[ggid[name-'a'+1]] = nameggid[name-'a'+1] to get idgglist[id] to get name(the same as mm's..)*/for (int i = 1; i <= n; ++i) {scanf(" %c", gglist + i);ggid[gglist[i] - 'a' + 1]  = i;}for (int i = 1; i <= n; ++i) {scanf(" %c", mmlist + i);mmid[mmlist[i] - 'A' + 1] = i;}for (int i = 1; i <= n; ++i) {char ch;scanf(" %c%*c", &ch);for (int j = 1; j <= n; ++j) {gg[ch-'a'+1].push_back(getchar()-'A'+1);}}for (int i = 1; i <= n; ++i) {char ch;scanf(" %c%*c", &ch);for (int j = 1; j <= n; ++j) {mm[ch-'A'+1].push_back(getchar()-'a'+1);}}solve();if (T) putchar('\n');}return 0;}

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