CF-weekly4 D. Haar Features

https://codeforces.com/gym/253910/problem/D

D. Haar Features

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The first algorithm for detecting a face on the image working in realtime was developed by Paul Viola and Michael Jones in 2001. A part of the algorithm is a procedure that computes Haar features. As part of this task, we consider a simplified model of this concept.

Let's consider a rectangular image that is represented with a table of size n × m. The table elements are integers that specify the brightness of each pixel in the image.

A feature also is a rectangular table of size n × m. Each cell of a feature is painted black or white.

To calculate the value of the given feature at the given image, you must perform the following steps. First the table of the feature is put over the table of the image (without rotations or reflections), thus each pixel is entirely covered with either black or white cell. The value of a feature in the image is the value of W - B, where W is the total brightness of the pixels in the image, covered with white feature cells, and Bis the total brightness of the pixels covered with black feature cells.

Some examples of the most popular Haar features are given below.

Your task is to determine the number of operations that are required to calculate the feature by using the so-called prefix rectangles.

A prefix rectangle is any rectangle on the image, the upper left corner of which coincides with the upper left corner of the image.

You have a variable value, whose value is initially zero. In one operation you can count the sum of pixel values ​​at any prefix rectangle, multiply it by any integer and add to variable value.

You are given a feature. It is necessary to calculate the minimum number of operations required to calculate the values of this attribute at an arbitrary image. For a better understanding of the statement, read the explanation of the first sample.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns in the feature.

Next n lines contain the description of the feature. Each line consists of m characters, the j-th character of the i-th line equals to "W", if this element of the feature is white and "B" if it is black.

Output

Print a single number — the minimum number of operations that you need to make to calculate the value of the feature.

Examples

input

Copy

6 8
BBBBBBBB
BBBBBBBB
BBBBBBBB
WWWWWWWW
WWWWWWWW
WWWWWWWW

output

Copy

2

input

Copy

3 3
WBW
BWW
WWW

output

Copy

4

input

Copy

3 6
WWBBWW
WWBBWW
WWBBWW

output

Copy

3

input

Copy

4 4
BBBB
BBBB
BBBB
BBBW

output

Copy

4

Note

The first sample corresponds to feature B, the one shown in the picture. The value of this feature in an image of size 6 × 8 equals to the difference of the total brightness of the pixels in the lower and upper half of the image. To calculate its value, perform the following two operations:

1. add the sum of pixels in the prefix rectangle with the lower right corner in the 6-th row and 8-th column with coefficient 1 to the variable value (the rectangle is indicated by a red frame);
2. add the number of pixels in the prefix rectangle with the lower right corner in the 3-rd row and 8-th column with coefficient  - 2 and variable value.

Thus, all the pixels in the lower three rows of the image will be included with factor 1, and all pixels in the upper three rows of the image will be included with factor 1 - 2 =  - 1, as required.

 

题目大意：

给出一个n*m的棋盘，每个小方格都是白色或者黑色，要计算白色格子的个数减去黑色格子个数的结果

给出一个操作：每次操作可以选择一个方格，然后计算它的前缀方格的个数乘以一个系数的结果。

问要得到最终答案，至少进行几次操作

题解：

从后往前枚举，即n->1,m->1，每当发现一个白色的格子不是1，或者黑色的格子不是-1，就将它以及它的前缀矩阵加上相应的值，使当前格子的值变成相应的1或者-1

复杂度为O(n^4)

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 110
using namespace std;
int n,m,a[maxn][maxn],ans;
char s[maxn][maxn];
int main(){
//    freopen("Cola.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]+1);
    for(int i=n;i>=1;i--){
        for(int j=m;j>=1;j--){
            if(s[i][j]=='W'&&a[i][j]!=1){
                int add=1-a[i][j];
                ans++;
                for(int x=i;x>=1;x--)
                    for(int y=j;y>=1;y--)
                        a[x][y]+=add;
            }
            else if(s[i][j]=='B'&&a[i][j]!=-1){
                int add=-1-a[i][j];
                ans++;
                for(int x=i;x>=1;x--)
                    for(int y=j;y>=1;y--)
                        a[x][y]+=add;
            }
        }
    }
    printf("%d\n",ans);
    return 0;
}