poj 2226(二分图最小顶点覆盖)

poj 2226
题意：用宽度为1长度不限的木板将 * 盖住而不盖住 .。

我们考虑二分图中一条边为一个点。我们对横竖进行遍历，看只横和只竖所需要的木板并且标号，例如样例

4 4
*.*.
.***
***.
..*.

横着：（图中数字表示用的是第几块木板）

1.2.
.333
444.
..5.

竖着：

1.4.
.345
234.
..4.

我们发现每一个点都可以被两块方向不同的木板覆盖，所以就以这个为依据建边，边相当于一个点，那么只要求出最小顶点覆盖即可。

#include<cstdio> 
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
int r, c;
char map[100][100];
int hang[100][100], lie[100][100], sz_hang = 0, sz_lie = 0;
vector<int> G[3000];
int lk[3000], vis[3000], cnt;
bool hungary(int u) {
    for (int i = 0; i < (int)G[u].size(); i++) {
        int v = G[u][i];
        if (vis[v] != cnt) {
            vis[v] = cnt;
            if (!lk[v] || hungary(lk[v])) {
                lk[v] = u;
                return true;
            }
        }
    }
    return false;
}
void clean() {
    ms(lk, 0), ms(vis, 0), ms(hang, 0), ms(lie, 0);
}
int solve() {
    clean();
    for (int i = 1; i <= c; i++) map[0][i] = '\0';
    for (int i = 1; i <= r; i++) scanf("%s", map[i] + 1), map[i][0] = '\0';

    for (int i = 1; i <= r; i++) {
        for (int j = 1; j <= c; j++) {
            if (map[i][j] == '*' && map[i][j - 1] != '*') hang[i][j] = ++sz_hang;
            if (map[i][j] == '*' && map[i][j - 1] == '*') hang[i][j] = hang[i][j - 1];
        }
    }
    for (int i = 1; i <= c; i++) {
        for (int j = 1; j <= r; j++) {
            if (map[j][i] == '*' && map[j - 1][i] != '*') lie[j][i] = ++sz_lie;
            if (map[j][i] == '*' && map[j - 1][i] == '*') lie[j][i] = lie[j - 1][i];
        }
    }

    for (int i = 1; i <= r; i++) 
    for (int j = 1; j <= c; j++) G[hang[i][j]].push_back(lie[i][j]);

    int ans = 0;
    for (int i = 1; i <= sz_hang; i++) ans += hungary(cnt = i);

    printf("%d", ans);
    return 0; 
}
int main() {
    scanf("%d%d", &r, &c), solve();
    return 0;
}