第二周上二分图（Oil Skimming HDU - 4185）

浮油采集算法

最新推荐文章于 2021-01-16 21:00:13 发布

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本文介绍了一种通过构建二分图并使用匹配算法来解决最大浮油采集次数的问题。该问题设定为在一个N×N的网格中，部分单元格被浮油覆盖，目标是计算能够连续采集两次浮油的最大次数。

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

题意：
海面上有一些浮油，现在要去打捞这些浮油，前提是一次要打捞连着的两块,求最多能打捞几次；

题解：
先遍历整个图，对每个油田进行编号，然后对于每个油田，和它相连的油田之间建立一条路，之后用二分图，求解

代码：

#include"stdio.h"
#include"string.h"
#include"algorithm"
using namespace std;
char map[1010][1010];
int e[1010][1010],mapp[2500][2500],book1[1010],book2[1010];
int cnt,n,to[4][2]={0,1,0,-1,1,0,-1,0};
bool dfs(int x)
{
	int i;
	for(i=0;i<cnt;i++)
	{
		if(mapp[x][i]&&!book2[i])
		{
			book2[i]=1;
			if(book1[i]==-1||dfs(book1[i]))//如果这个油田还没被选或者之前选这个油田的油田可以找到另一个油田相连
			{
				book1[i]=x;
				return 1;
			}
		}
	}
	return 0;
}
int find()
{
	int i,sum=0;
	memset(book1,-1,sizeof(book1));
	for(i=0;i<cnt;i++)
	{
		memset(book2,0,sizeof(book2));
		if(dfs(i))//如果找到了
		sum++;
	}
	return sum;
}
int main()
{
	int t,h=1;
	scanf("%d",&t);
	while(t--)
	{
		memset(mapp,0,sizeof(mapp));
		memset(e,-1,sizeof(e));
		scanf("%d",&n);
		int i,j,k;
		cnt=0;
		for(i=0;i<n;i++)
		scanf("%s",map[i]);
		for(i=0;i<n;i++)//对每个油田进行编号
		{
			for(j=0;j<n;j++)
			{
				if(map[i][j]=='#')
				e[i][j]=cnt++;
			}
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				if(map[i][j]=='#')//找到第一个油田之后找他四周的点
				{
					for(k=0;k<4;k++)
					{
						int tx=i+to[k][0];
						int ty=j+to[k][1];
						if(map[tx][ty]=='#')//有相连的油田，建立连接
						mapp[e[i][j]][e[tx][ty]]=1;
					}
				}
			}
		}
		int s=find();//进行二分查找
		printf("Case %d: %d\n",h++,s/2);//由于有重复的情况，所以结果除以2
	}
	return 0;
}