POJ 1719 Shooting Contest(二分图匹配)

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POJ 1719 Shooting Contest

题目链接

题意:给定一个矩阵,每列有两个白点,其他都是黑点,现在要求每列选一个白点,使得每一行至少包含一个白点被选中

思路:二分图匹配,白点的位置行列建边,然后跑匹配,如果匹配数不等于行数,就是是无解,然后输出方案的时候注意,如果有位置是没匹配的,说明这个位置是多余的,但是还是要任意选一个白点来输出

代码:

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int N = 1005;int t, r, c;vector<int> g[N];int left[N], vis[N];bool dfs(int u) {for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (vis[v]) continue;vis[v] = 1;if (!left[v] || dfs(left[v])) {left[v] = u;return true;}}return false;}int save[N];void hungary() {int ans = 0;memset(left, 0, sizeof(left));for (int i = 1; i <= r; i++) {memset(vis, 0, sizeof(vis));if (dfs(i)) ans++;}if (ans != r) {printf("NO\n");return;}for (int i = 1; i <= c; i++)printf("%d%c", left[i] ? left[i] : save[i], i == c ? '\n' : ' ');}int main() {scanf("%d", &t);while (t--) {scanf("%d%d", &r, &c);for (int i = 1; i <= r; i++)g[i].clear();int tmp;for (int i = 1; i <= c; i++) {for (int j = 0; j < 2; j++) {scanf("%d", &tmp);save[i] = tmp;g[tmp].push_back(i);}}hungary();}return 0;}


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