POJ 2226 Muddy Fields（最小点覆盖）

题目链接：
POJ 2226 Muddy Fields
题意：
和POJ 3041 类似。只不过每次只可以清除一列或一行中连续的点，问最少的清除次数。
分析：
不再将每行和每列看成二分图的点，而是将一行中连续的点和一列中连续的点看成二分图的点，建好图下面就是套路了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#include <map>
#include <string>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 3000;

int n, m, total, cntx, cnty;
int head[MAX_N], vis[MAX_N], match[MAX_N], x[MAX_N][MAX_N], y[MAX_N][MAX_N];
char s[MAX_N][MAX_N];

struct Edge{
    int to, next;

    Edge() {}
    Edge(int _to, int _next) : to(_to), next(_next){
    }
}edge[MAX_N*MAX_N];

inline void AddEdge(int from, int to)
{
    edge[total].to = to;
    edge[total].next = head[from];
    head[from] = total++;
}

inline bool dfs(int u)
{
    for(int i = head[u]; i != -1; i = edge[i].next){
        int to = edge[i].to;
        if(!vis[to]){
            vis[to] = 1;
            if(match[to] == -1 || dfs(match[to])){
                match[to] = u;
                return true;
            }
        }
    }
    return false;
}

inline void hungary()
{
    int ans = 0;
    for(int i = 0; i < cntx; ++ i){
        memset(vis,0,sizeof(vis));
        if(dfs(i)) ans++;
    }
    cout << ans << endl;
}

int main()
{
    freopen("E.in", "r", stdin);
    IOS;
    while(cin >> n >> m){
        memset(head, -1, sizeof(head));
        memset(match, -1, sizeof(match));
        cntx = cnty = total = 0;

        for(int i = 0; i < n; ++ i){
            cin >> s[i];
        }

        for(int i = 0; i < n; i++){
            for(int j = 0; j < m;){
                if(s[i][j] == '*'){
                    int end = m-1;
                    for(int k = j; k < m; k++){
                        if( s[i][k] == '.'){
                            end = k-1;
                            break;
                        }
                    }
                    for(int k = j; k <= end; k++){
                        x[i][k] = cntx;
                    }
                    cntx++;
                    j = end + 1;
                }else {
                    j++;
                }
            }
        }

        for(int j = 0; j < m; j++){
            for(int i = 0; i < n;){
                if(s[i][j] == '*'){
                    int end = n-1;
                    for(int k = i; k < n; k++){
                        if( s[k][j] == '.'){
                            end = k-1;
                            break;
                        }
                    }
                    for(int k = i; k <= end; k++){
                        y[k][j] = cnty;
                    }
                    cnty++;
                    i = end + 1;
                }else {
                    i++;
                }
            }
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(s[i][j] == '*'){
                    AddEdge(x[i][j], y[i][j]);
                }
            }
        }
        hungary();
    }
    return 0;
}