POJ

题意

农夫的养牛场，是一个R 行C 列的矩形，一场大雨后，养牛场低洼的地方都有了积水。农夫的牛都很娇贵的，他们吃草的时候，不想把他们的蹄子给弄脏了。为了不让牛儿们把它们的蹄子弄脏，农夫决定把有水的地方铺上木板。他的木板是宽度为1，长度没有限制的。
他想用最少数目的木板把所有有水的低洼处给覆盖上，木板不能覆盖草地，但是可以重叠。

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题解

找到每行连续的’*’，对每一块编号。再找到每列连续的，编号。每行的和每列的连边跑匈牙利即可

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常数巨大的丑陋代码

# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# include <algorithm>
using namespace std;

# define IL inline
# define RG register
# define UN unsigned
# define ll long long
# define rep(i, a, b) for(RG int i = a; i <= b; i++)
# define per(i, a, b) for(RG int i = b; i >= a; i--)
# define mem(a, b) memset(a, b, sizeof(a))
# define max(a, b) (((a) > (b)) ? (a) : (b))
# define min(a, b) (((a) < (b)) ? (a) : (b))
# define Swap(a, b) a ^= b, b ^= a, a ^= b;

IL ll Get(){
    char c = '!'; ll z = 1, num = 0;
    while(c != '-' && (c < '0' || c > '9')) c = getchar();
    if(c == '-') z = -1, c = getchar();
    while(c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
    return num * z;
}

const int MAXN = 501;
int map[MAXN][MAXN], vis[MAXN], girl[MAXN], n, m, m1[MAXN][MAXN], m2[MAXN][MAXN];
char land[MAXN][MAXN];

IL int Dfs(RG int u){
    rep(v, 1, m) if(map[u][v] && !vis[v]){
        vis[v] = 1;
        if(girl[v] == -1 || Dfs(girl[v])){
            girl[v] = u;
            return 1;
        }
    }
    return 0;
}

IL int Hungarian(){
    RG int ans = 0;
    mem(girl, -1);
    rep(i, 1, n) mem(vis, 0), ans += Dfs(i);
    return ans;
}

int main(){
    RG int x = Get(), y = Get();
    rep(i, 1, x) rep(j, 1, y) scanf(" %c", &land[i][j]);
    rep(i, 1, x) rep(j, 1, y) if(land[i][j] == '*'){
        n++;
        while(j <= y && land[i][j] == '*') m1[i][j++] = n;
    }
    rep(i, 1, y) rep(j, 1, x) if(land[j][i] == '*'){
        m++;
        while(j <= x && land[j][i] == '*') m2[j++][i] = m;
    }
    rep(i, 1, x) rep(j, 1, y) if(land[i][j] == '*') map[m1[i][j]][m2[i][j]] = 1;
    printf("%d
", Hungarian());
    return 0;
}