POJ 1548 Robots（最小路径覆盖）

POJ 1548 Robots

题目链接

题意：乍一看还以为是小白上那题dp，事实上不是，就是求一共几个机器人能够覆盖全部路径

思路：最小路径覆盖问题。一个点假设在还有一个点右下方，就建边。然后跑最小路径覆盖就可以

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 25 * 25;

int x[N], y[N], cnt = 1;
vector<int> g[N];

bool judge(int i, int j) {
	return x[j] >= x[i] && y[j] >= y[i];
}

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] == -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 < cnt; i++) {
		memset(vis, 0, sizeof(vis));
		if (dfs(i)) ans++;
	}
	return ans;
}

int main() {
	while (~scanf("%d%d", &x[0], &y[0])) {
		if (x[0] == -1 && y[0] == -1) break;
		while (~scanf("%d%d", &x[cnt], &y[cnt])) {
			if (x[cnt] == 0 && y[cnt] == 0) break;
			cnt++;
		}
		for (int i = 0; i < cnt; i++) {
			g[i].clear();
			for (int j = 0; j < i; j++) {
				if (judge(i, j)) g[i].push_back(j);
				if (judge(j, i)) g[j].push_back(i);
			}
		}
		printf("%d\n", cnt - hungary());
		cnt = 1;
	}
	return 0;
}