Codeforces Round #139 (Div. 2)C. Barcode

C. Barcode

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

11

input

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

output

5

題意：給你一個由兩種顏色組成的圖，讓你進行染色，要求染色完成的圖滿足每列顏色相同，每行的顏色至少連續x種顏色的相同且連續相同顏色數小於等於

      y，求最少的染色次數。

思路：先統計每列顏色爲"#"或"."的染色次數，用dp1[i][1]表示前i列滿足條件(且第i列爲“#”，dp[i][0]對應)的最少染色數，

dp1[i][1]=min(dp1[i][1],dp1[i-k][0]+dp[i][0]-dp[i-k][0]),dp1[i]表示第i列每列顏色滿足要求的染色次數.

代碼：

<pre name="code" class="cpp">#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define inf 9999999

char ma[1005][1005];
int dp[1005][2],dp1[1005][2];
int main()
{
    int n,m,x,y;
    while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%s",ma[i]+1);
        }
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(ma[j][i]=='#')
                    dp[i][1]++;
                else
                    dp[i][0]++;
            }
        }
        dp[0][0]=0;
        dp[0][1]=0;
        for(int i=1;i<=m;i++)
        {
            dp[i][0]=dp[i-1][0]+dp[i][0];//統計前i列‘.’的個數
            dp[i][1]=dp[i-1][1]+dp[i][1];//統計前i列'#'的個數
        }
        for(int i=1;i<=m;i++)
        {
            dp1[i][0]=inf;
            dp1[i][1]=inf;
        }
        dp1[0][0]=0;
        dp1[0][1]=0;
        for(int i=1;i<=m;i++)
        {
            for(int k=x;k<=y;k++)
            {
              if(i-k>=0)
              {
                  dp1[i][0]=min(dp1[i-k][1]+dp[i][1]-dp[i-k][1],dp1[i][0]);//第i列爲‘.’且滿足要求的最小染色次數
                  dp1[i][1]=min(dp1[i-k][0]+dp[i][0]-dp[i-k][0],dp1[i][1]);
              }
            }
        }
        printf("%d\n",min(dp1[m][1],dp1[m][0]));
    }
    return 0;
}