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;

}