题意：在一个n*m的草地上，.代表草地，*代表水，现在要用宽度为1，长度不限的木板盖住水，木板可以重叠，但是所有的草地都不能被木板覆盖。问至少需要的木板数。
分析：“2要素”经典题，详见《算法竞赛进阶指南》P429。
代码：
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 56;
int n, m, tot = 1, a[N][N][2], f[N*N], ans;
char s[N][N];
bool v[N*N];
vector<int> e[N*N];bool dfs(int x) {for (unsigned int i = 0; i < e[x].size(); i++) {int y = e[x][i];if (v[y]) continue;v[y] = 1;if (!f[y] || dfs(f[y])) {f[y] = x;return 1;}}return 0;
}int main() {cin >> n >> m;for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);for (int i = 1; i <= n; i++)for (int j = 1; j <= m + 1; j++)if (s[i][j] == '*') a[i][j][0] = tot;else ++tot;int t = tot;for (int j = 1; j <= m; j++)for (int i = 1; i <= n + 1; i++)if (s[i][j] == '*') a[i][j][1] = tot;else ++tot;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)if (s[i][j] == '*') {e[a[i][j][0]].push_back(a[i][j][1]);e[a[i][j][1]].push_back(a[i][j][0]);}for (int i = 1; i < t; i++) {memset(v, 0, sizeof(v));ans += dfs(i);}cout << ans << endl;return 0;
}