Painter's Problem POJ1681

最新推荐文章于 2020-01-27 10:07:07 发布

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最新推荐文章于 2020-01-27 10:07:07 发布

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分类专栏：状态压缩算法学习文章标签： express integer character output each input

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本文探讨了一种解决在给定约束条件下最小化刷墙次数的问题，通过位运算和状态枚举的方法来优化算法效率，确保所有墙面最终变为黄色。

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Painter's Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2917		Accepted: 1450

Description

There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob's brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow.

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a 'w' to express a white brick while a 'y' to express a yellow brick.

Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can't paint all the bricks yellow, print 'inf'.

Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww

Sample Output

0
15

题意：

刷墙每次刷一格会将上下左右中五个格子变色，求最少的刷方法使得所有的格子都变成yellow。

分析：

与POJ1185 炮兵阵地有相似之处，状态枚举和位运算。同样从第一行开始枚举所有粉刷状态，然后后面每行每格paint[k][p]的粉刷情况要根据wall[k-1][p]经过paint[k-1][p]、paint[k-1][p-1]、paint[k-1][p+1]、paint[k-2][p]后的情况而定。若为1（即白色）则paint[k][p]=1。所有行处理完后，判断墙壁是否已经全为黄色，并记录下paint[i][j]为1的个数。循环至第一行枚举了0000 - 1111所有状态。

程序源代码如下：

#include <iostream>
using namespace std;

bool wall[18][18];
bool paint[18][18];
int count=0;
int mmin=300;

bool allYellow(int n)
{
	int i,j;
	count=0;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			if(wall[i][j]^paint[i][j]^paint[i][j-1]^paint[i][j+1]^paint[i-1][j]^paint[i+1][j])return false;
			if(paint[i][j])count++;
		}
	}
	return true;
}

int main()
{
	freopen("in.txt","r",stdin);
	int t,n;
	cin>>t;
	int i,j,k,p;
	char wy[16];
	int mmax,tmp;
	bool possible=false;

	for(i=0;i<t;i++)
	{
		cin>>n;	
		memset(wall,0,sizeof(wall));

		for(j=1;j<=n;j++)
		{
			cin>>wy;
			for(k=0;k<n;k++)
			{
				if(wy[k]=='w')wall[j][k+1]=true;
			}
		}

		mmax=1<<n;
		mmin=300;
		possible=false;

		for(j=0;j<mmax;j++)			//第一行的状态 0000 - 1111 枚举
		{
			memset(paint,0,sizeof(paint));
			tmp=j;
			for(k=1;k<=n;k++)      //第一行的状态
			{
				paint[1][k]=tmp&1;
				tmp=tmp>>1;       //右移一位
			}

			for(k=2;k<=n;k++)     
			{
				for(p=1;p<=n;p++)		//根据wall[k-1][p]的paint后状态 决定paint[k][p]值
				{
					if(wall[k-1][p]^paint[k-1][p]^paint[k-1][p-1]^paint[k-1][p+1]^paint[k-2][p])paint[k][p]=true;
				}
			}

			if(allYellow(n))
			{
				possible=true;
				if(mmin>count)mmin=count;
			}

		}
		if(possible)cout<<mmin<<endl;
		else cout<<"inf"<<endl;
	}
	return 0;
}