Codeforce 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

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:

.#.#.
.#.#.

分析：先预处理出从第一列到i列的白点与黑点的数量。设dp[i][j]为前i列的以第i列结尾的颜色为j的合法状态下的最小更改数量，则可递推出dp[i+x][]...dp[i+y][]。

Code：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#define eps 1e-7
#define LL long long
#define pb push_back
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;

const int inf=0x3f3f3f3f;
int B[1005],W[1005],dp[1005][2];
char g[1005][1005];
int n,m,x,y;

int main()
{
scanf("%d %d %d %d",&n,&m,&x,&y);
for(int i=1;i<=n;i++) scanf("%s",g[i]+1);
memset(B,0,sizeof(B));
memset(W,0,sizeof(W));
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(g[j][i]=='#') B[i]++;
else W[i]++;
}
B[i]+=B[i-1]; W[i]+=W[i-1];
}
memset(dp,inf,sizeof(dp));
dp[0][0]=dp[0][1]=0;
for(int i=0;i<m;i++){
for(int j=x;j<=y&&i+j<=m;j++){
dp[i+j][0]=Min(dp[i+j][0],dp[i][1]+B[i+j]-B[i]);
dp[i+j][1]=Min(dp[i+j][1],dp[i][0]+W[i+j]-W[i]);
}
}
printf("%d\n",Min(dp[m][0],dp[m][1]));
return 0;
}