CodeForces 225C. Barcode

最新推荐文章于 2020-12-05 10:24:09 发布

陌初

最新推荐文章于 2020-12-05 10:24:09 发布

阅读量719

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划文章标签： dp codeforces

本文链接：https://blog.youkuaiyun.com/qq_25931695/article/details/43730413

动态规划专栏收录该内容

8 篇文章

订阅专栏

本文深入探讨了代码优化的策略与算法在实际开发中的应用，从基础优化技巧到高级算法选择，全面覆盖，旨在提升开发效率与代码质量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接： http://codeforces.com/problemset/problem/225/C

You've got an n × m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.

A picture is a barcode if the following conditions are fulfilled:

All pixels in each column are of the same color.
The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.

Input

The first line contains four space-separated integers n, m, x and y (1 ≤ n, m, x, y ≤ 1000; x ≤ y).

Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".

Output

In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.

Sample test(s)

Input

6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..

Output

Input

2 5 1 1
#####
.....

Output

Note

In the first test sample the picture after changing some colors can looks as follows:

.##..
.##..
.##..
.##..
.##..
.##..

In the second test sample the picture after changing some colors can looks as follows:

.#.#.
.#.#.

题意：

给出n行m列，其中 ‘#’ 为黑色，‘.’ 为白色，要求改变其中的符号，使得变成宽度为【x,y】的黑白相间的条子。

思路：

先统计#的个数，然后有两个状态方程

dp1[ i ]=max( dp2[ i - j ]+a[ i ]-a[ i - j ] , dp1[ i ] );//dp1表示涂成白的，式子表示【前 i-j 列涂黑，j列涂白】与【前 i 列都涂白】的最优方案

dp2[ i ]=max( dp1[ i - j ]+j*n-a[ i ]+a[ i - j ],dp2[ i ]);//dp1表示涂成黑的，式子表示【前 i-j 列涂白，j列涂黑】与【前 i 列都涂黑】的最优方案

代码如下：

#include <stdio.h>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 1005
int main()
{
	int n,m,x,y;
	int i,j,dp1[N],dp2[N],a[N];
	char ch;
	while(~scanf("%d%d%d%d",&n,&m,&x,&y))
	{
		memset(a,0,sizeof(a));
		for(i=1;i<=n;i++)
		{
		   getchar(); 
		   for(j=1;j<=m;j++)
		   {
   			      scanf("%c",&ch);
				  if(ch=='#')a[j]++;
   		    }
		}
		 a[0]=0;
   		 for(i=1;i<=m;i++)
			a[i]+=a[i-1]; 
		 memset(dp1,127,sizeof(dp1));
		 memset(dp2,127,sizeof(dp2));
   	         dp1[0]=dp2[0]=0;
		 for(i=1;i<=m;i++)
		    for(j=x ; j<=y && j<=i ; j++)
			{
				dp1[i]=min(dp2[i-j]+a[i]-a[i-j],dp1[i]);
				dp2[i]=min(dp1[i-j]+j*n-(a[i]-a[i-j]),dp2[i]);
			}	   
		 printf("%d\n",min(dp1[m],dp2[m]));	
	}
	return 0; 
}