POJ 2226 Muddy Fields

题目链接

题意：给定一个图，要求用纸片去覆盖‘*‘的位置，纸片可以重叠，但是不能放到‘.‘的位置，为最少需要几个纸片

思路：二分图匹配求最小点覆盖，和放车那题基本一样，就是注意要预处理一下行列，把连续横的‘*‘当成一行，竖的‘*‘当成一列，建图跑最小点覆盖即可

代码：

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

const int N = 55;
const int M = 1505;

int n, m, tox[N][N], toy[N][N], xn, yn;
char str[N][N];
vector<int> g[M];

int left[M], vis[M];

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 < xn; i++) {
		memset(vis, 0, sizeof(vis));
		if (dfs(i)) ans++;
	}
	return ans;
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		int cnt = 0;
		for (int i = 0; i < n; i++) {
			scanf("%s", str[i]);
			int flag = 0;
			for (int j = 0; j < m; j++) {
				if (str[i][j] == '*') {
					tox[i][j] = cnt;
					flag = 1;
				} else if (str[i][j] == '.' && flag) {
					g[cnt].clear();
					cnt++;
					flag = 0;
				}
			}
			if (flag) {
				g[cnt].clear();
				cnt++;
			}
			xn = cnt;
		}
		cnt = 0;
		for (int i = 0; i < m; i++) {
			int flag = 0;
			for (int j = 0; j < n; j++) {
				if (str[j][i] == '*') {
					toy[j][i] = cnt;
					flag = 1;
				} else if (str[j][i] == '.' && flag) {
					cnt++;
					flag = 0;
				}
			}
			if (flag) 
				cnt++;
			yn = cnt;
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (str[i][j] == '*') {
					g[tox[i][j]].push_back(toy[i][j]);
				}
			}
		}
		printf("%d\n", hungary());
	}
	return 0;
}

POJ 2226 Muddy Fields（最小点覆盖）