Codeforces 225C Barcode【dp】

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我是靠谱客的博主冷傲电话，最近开发中收集的这篇文章主要介绍Codeforces 225C Barcode【dp】，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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.

Examples

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:

.#.#.
.#.#.

题目大意：

给你一个N*M的矩阵，其中有白色和黑色两种颜色，我们可以将其中的任意格子颜色变成另外一种颜色，问最少操作多少次（变换颜色），使得这个N*M的矩阵变成一个条形码，并且保证相同颜色挨在一起的列数不小于x不大于y、

思路：

1、首先预处理出每一列变成一种颜色需要多少花费，记做a【i】【2】，其中a【i】【0】表示将第i列都变成点的花费，a【i】【1】表示将第i列都变成#的花费。

2、那么考虑dp过程：

①设定dp【i】【j】【k（0/1）】表示现在进行到第i列上，当前列颜色为k（0/1），并且其前边的列与当前列颜色相同的连续列数为j的最小花费。

②那么不难推出其状态转移方程：
if（j==1）dp【i】【j】【k】=min（dp【i-1】【l】【k】，dp【i】【j】【k】+a【i】【k】）【x<=l<=y】

表示当前列的颜色要与上一列的颜色不同，那么其要满足上一列的颜色连续的列数在x到y之间，在这个区间内，维护最小值。

if（j！=1）dp【i】【j】【k】=dp【i-1】【j-1】【k】+a【i】【k】；

表示当前列的颜色要与上一列的颜色相同，那么直接从上一列的状态转移过来即可。

③时间复杂度O（m^2）

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char map[1005][1005];
int dp[1005][1005][2];
int a[1005][2];
int main()
{
int n,m,x,y;
while(~scanf("%d%d%d%d",&n,&m,&x,&y))
{
memset(dp,0x3f3f3f3f,sizeof(dp));
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
}
//a[j][0] 都弄成#
//a[j][1] 都弄成.
for(int j=0;j<m;j++)
{
int cnt=0;
for(int i=0;i<n;i++)
{
if(map[i][j]=='#')cnt++;
}
a[j][0]=cnt;
a[j][1]=n-cnt;
}
for(int i=0;i<m;i++)
{
if(i==0)
{
dp[i][1][0]=a[i][0];
dp[i][1][1]=a[i][1];
continue;
}
for(int j=1;j<=i+1&&j<=y;j++)
{
if(j==1)
{
for(int k=x;k<=y;k++)
{
dp[i][j][0]=min(dp[i][j][0],dp[i-1][k][1]+a[i][0]);
dp[i][j][1]=min(dp[i][j][1],dp[i-1][k][0]+a[i][1]);
}
}
else
{
dp[i][j][0]=dp[i-1][j-1][0]+a[i][0];
dp[i][j][1]=dp[i-1][j-1][1]+a[i][1];
}
}
}
int output=0x3f3f3f3f;
for(int i=x;i<=y;i++)
{
output=min(dp[m-1][i][0],output);
output=min(dp[m-1][i][1],output);
}
printf("%dn",output);
}
}