POJ 2226 Muddy Fields（最小顶点覆盖）

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。纸片能够重叠。可是不能放到'.'的位置，为最少须要几个纸片

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，建图跑最小点覆盖就可以

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

, toy

, xn, yn;
char str

;
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;
}