HDU 2236 无题II（二分图匹配+二分）

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HDU 2236 无题II

题目链接

思路：行列只能一个，想到二分图，然后二分区间长度，枚举下限，就能求出哪些边是能用的，然后建图跑二分图，如果最大匹配等于n就是符合的

代码：

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int N = 105;int t, n, x[N][N], have[N], hn;int vis[N], left[N];vector<int> g[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] == -1 || dfs(left[v])) {left[v] = u;return true;}}return false;}int hungary() {int ans = 0;memset(left, -1, sizeof(left));for (int i = 0; i < n; i++) {memset(vis, 0, sizeof(vis));if (dfs(i)) ans++;}return ans;}bool judge(int len) {for (int i = 0; i < hn; i++) {for (int j = 0; j < n; j++) g[j].clear();int down = have[i], up = have[i] + len;for (int u = 0; u < n; u++)for (int v = 0; v < n; v++)if (x[u][v] >= down && x[u][v] <= up)g[u].push_back(v);if (hungary() == n) return true;}return false;}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);memset(vis, 0, sizeof(vis));hn = 0;for (int i = 0; i < n; i++)for (int j = 0; j < n; j++) {scanf("%d", &x[i][j]);if (vis[x[i][j]]) continue;vis[x[i][j]] = 1;have[hn++] = x[i][j];}int l = 0, r = 101;while (l < r) {int mid = (l + r) / 2;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", l);}return 0;}

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