首页 > ACM题库 > HDU-杭电 > HDU 1801 Annoying painting tool-贪心-[解题报告] C++
2013
12-23

HDU 1801 Annoying painting tool-贪心-[解题报告] C++

Annoying painting tool

问题描述 :

Maybe you wonder what an annoying painting tool is? First of all, the painting tool we speak of supports only black and white. Therefore, a picture consists of a rectangular area of pixels, which are either black or white. Second, there is only one operation how to change the colour of pixels:

Select a rectangular area of r rows and c columns of pixels, which is completely inside the picture. As a result of the operation, each pixel inside the selected rectangle changes its colour (from black to white, or from white to black).

Initially, all pixels are white. To create a picture, the operation described above can be applied several times. Can you paint a certain picture which you have in mind?

输入:

The input contains several test cases. Each test case starts with one line containing four integers n, m, r and c. (1 ≤ r ≤ n ≤ 100, 1 ≤ c ≤ m ≤ 100), The following n lines each describe one row of pixels of the painting you want to create. The ith line consists of m characters describing the desired pixel values of the ith row in the finished painting (’0′ indicates white, ’1′ indicates black).

The last test case is followed by a line containing four zeros.

输出:

For each test case, print the minimum number of operations needed to create the painting, or -1 if it is impossible.

样例输入:

3 3 1 1
010
101
010
4 3 2 1
011
110
011
110
3 4 2 2
0110
0111
0000
0 0 0 0

样例输出:

4
6
-1

题目链接:

题目大意:该题是利用贪心算法,自左向右,自上而下遍历一次,遇到 与所给的颜色不同,即改变一次,最后判断与所给的是否完全相同

代码如下:

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=110;
int map[maxn][maxn],color[maxn][maxn];
char s[maxn][maxn];
int n,m,r,c,ans;
void change()
{
	int i,j,k,p;
	for(i=0;i<n;i++)
	{
		for(j=0;j<m;j++)
		{
			if(i+r<=n&&j+c<=m&&map[i][j]!=color[i][j])//限制寻找条件,在边界范围之内,并且与所给的颜色不同
			{
				for(k=i;k<i+r;k++)
				{
					for(p=j;p<j+c;p++)
						if(color[k][p])
							color[k][p]=0;
						else
							color[k][p]=1;
				}
				ans++;
			}
			
		}
	}
}
int judge()//判断最后是否与所给的颜色完全相同
{
	int i,j,k;
	for(i=0;i<n;i++)
	{
		for(j=0;j<m;j++)
		{
			if(map[i][j]!=color[i][j])
				return 0;
		}
	}
	return 1;
}

int main(void)
{
	int i,j;
	while(scanf("%d%d%d%d",&n,&m,&r,&c)!=EOF)
	{
		if(n==0&&m==0&&r==0&&c==0)
			break;
		memset(map,0,sizeof(map));//对其进行初始化
		memset(color,0,sizeof(color));
		memset(s,0,sizeof(s));
		ans=0;
		for(i=0;i<n;i++)
		{
			scanf("%s",s[i]);
			for(j=0;j<m;j++)
			{
				map[i][j]=s[i][j]-'0';
			}
		}
		change();
		if(judge())
			printf("%d\n",ans);
		else
			printf("-1\n");
	}
	return 0;
}

 

解题报告转自:http://blog.csdn.net/chao_xun/article/details/8039413


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