Hdu4185 Oil Skimming

Oil Skimming

Problem Description

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

 

Source

The 2011 South Pacific Programming Contest

 

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题目的意思是给出一张图#表示油田，用1*2的矩阵去盖在#上，问最多可以放多少个#

思路：题目就是拿行列和为奇数的点和行列和为偶的点匹配，问最多可以多少，图大hash一下

其实复杂度还是绷不住，我也没什么思路，但看了网上的都是直接上，数据水了

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int MAXN=1005;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];
int link[MAXN];
int ha[MAXN][MAXN];
char s[MAXN];
int dir[4][2]= {{-1,0},{1,0},{0,-1},{0,1}};

bool dfs(int u)
{
    int v;
    for(v=0; v<vN; v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0; u<uN; u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}

int main()
{
    int m,n,k,x,y,T;
    int q=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);

        int cnt=0;
        memset(ha,-1,sizeof ha);
        for(int i=0; i<n; i++)
        {
            scanf("%s",s);
            for(int j=0; j<n; j++)
            {
                if(s[j]=='#')
                    ha[i][j]=cnt++;
            }
        }
        memset(g,0,sizeof g);
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                if(ha[i][j]!=-1)
                {
                    for(int k=0; k<4; k++)
                    {
                        int xx=i+dir[k][0];
                        int yy=j+dir[k][1];
                        if(xx>=0&&xx<n&&yy>=0&&yy<n&&ha[xx][yy]!=-1)
                        {
                            g[ha[i][j]][ha[xx][yy]]=1;
                        }
                    }

                }
        uN=vN=cnt;
        printf("Case %d: %d\n",q++,hungary()/2);

    }
    return 0;
}