HDU 2236 无题II（二分图匹配+二分）

Posted 2020-10-28 zhchoutai

HDU 2236 无题II

题目链接

思路：行列仅仅能一个，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，假设最大匹配等于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;
}