G - Oil Skimming ——二分图匹配（匈牙利算法）

G - Oil Skimming

Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such oil baron has a special plane that can skim the surface of the water collecting oil on the water's surface. However, each scoop covers a 10m by 20m rectangle (going either east/west or north/south). It also requires that the rectangle be completely covered in oil, otherwise the product is contaminated by pure ocean water and thus unprofitable! Given a map of an oil slick, the oil baron would like you to compute the maximum number of scoops that may be extracted. The map is an NxN grid where each cell represents a 10m square of water, and each cell is marked as either being covered in oil or pure water.

Input

The input starts with an integer K (1 <= K <= 100) indicating the number of cases. Each case starts with an integer N (1 <= N <= 600) indicating the size of the square grid. Each of the following N lines contains N characters that represent the cells of a row in the grid. A character of '#' represents an oily cell, and a character of '.' represents a pure water cell.

Output

For each case, one line should be produced, formatted exactly as follows: "Case X: M" where X is the case number (starting from 1) and M is the maximum number of scoops of oil that may be extracted.

Sample Input

1
6
......
.##...
.##...
....#.
....##
......

Sample Output

Case 1: 3

思路：

       是把 ‘#’  替换成 数字，构建 关系G(本代码是 line[][]) 。

注意：

        1）二分匹配时，要记得把n换成tmp。  详细见代码

        2）记得清零；

总结：

       一个二分图中的最大匹配数等于这个图中的 最小点覆盖数——假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖所有的边。

代码：

#include<map>
#include<queue>
#include<math.h>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define MM 605
const int INF=1e9;
int line[MM][MM],book[MM],man[MM],t[MM][MM];
char tu[MM][MM];
int n;
bool Find(int x)
{
    for(int i=1; i<=n; i++)
    {
        if(line[x][i] && !book[i])
        {
            book[i]=1;
            if(!man[i] || Find(man[i]))
            {
                man[i]=x;
                return true;
            }
        }
    }
    return false;
}
int match()
{
    int ans=0;
    for(int i=1; i<=n; i++)
    {
        mem(book,0);
        if(Find(i))
            ans++;
    }
    return ans/2;
}
int main()
{
    int ca,cas=1;
    scanf("%d",&ca);
    while(ca--)
    {
        mem(tu,'.');
        mem(t,0);
        mem(line,0);
        mem(man,0);
        int tmp=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%s",tu[i]+1);
            for(int j=1; j<=n; j++)
            {
                if(tu[i][j]=='#')
                    t[i][j]=++tmp;
            }
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(tu[i][j]=='#')
                {
                    if(tu[i-1][j]=='#')
                        line[t[i][j]][t[i-1][j]]=1;
                    if(tu[i][j-1]=='#')
                        line[t[i][j]][t[i][j-1]]=1;
                    if(tu[i+1][j]=='#')
                        line[t[i][j]][t[i+1][j]]=1;
                    if(tu[i][j+1]=='#')
                        line[t[i][j]][t[i][j+1]]=1;
                }
            }
        }
        n=tmp;
        printf("Case %d: %d\n",cas++,match());
    }
    return 0;
}