CF1200D White Lines | 前缀和

传送门

Examples

input 1

4 2
BWWW
WBBW
WBBW
WWWB

output 1

4

input 2

3 1
BWB
WWB
BWB

output 2

2

input 3

5 3
BWBBB
BWBBB
BBBBB
BBBBB
WBBBW

output 3

2

input 4

2 2
BW
WB

output 4

4

input 5

2 1
WW
WW

output 5

4

Note

In the first example, Gildong can click the cell(2,2), then the working screen becomes:

BWWW
WWWW
WWWW
WWWB

Then there are four white lines — the 2-nd and 3-rd row, and the 2-nd and 3-rd column.

In the second example, clicking the cell (2,3)makes the 2-nd row a white line.

In the third example, both the 2-nd column and 5-th row become white lines by clicking the cell (3,2).

 

题意：有一个n*n的格子，由'B'和'W'组成，B代表黑色，W代表白色。现在有一个k*k的橡皮擦，你可以选一个地方(i,j)(1≤i≤n-k+1,1≤j≤n-k+1)点击，它会将(i',j')（i≤i'≤i+k-1,j≤j'≤j+k-1）区域全部变为白色，若一行（列）全是W全是W那么这一行（列）就是一条白色的线。现在问你擦一次之后最多能有多少条线。

题解：我们可以利用前缀和统计每一行每一列有多少个黑格子，前缀和为0表示这一行（列）本来就是一条白线，可以算出初始白线的数量。然后我们一行行一列列判断从这个点开始往右（下）k个变成白色之后会不会增加一条线，再用前缀和记录前n行（列）一共能加几条线。最后枚举每一个可以点击的点看这个区间能增加多少白线更新ans。

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e3 + 10;
char s[N][N];
int r[N][N],c[N][N],rr[N][N],cc[N][N];
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for (int i = 1; i <= n; i++)
        scanf("%s",s[i]+1);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            r[i][j] = r[i][j-1]+(s[i][j]=='B');
            c[i][j] = c[i][j-1]+(s[j][i]=='B');
        }
    }
    int tot = 0;
    for (int i = 1; i <= n; i++) 
        tot+=(r[i][n] == 0) + (c[i][n] == 0);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n-k+1; j++) {
            rr[i][j] = rr[i-1][j] + (r[i][j+k-1] - r[i][j-1] == r[i][n] && r[i][n]);
            cc[i][j] = cc[i-1][j] + (c[i][j+k-1] - c[i][j-1] == c[i][n] && c[i][n]);
        }
    }
    int ans = tot;
    for (int i = 1; i <= n-k+1; i++) 
        for (int j = 1; j <= n-k+1; j++)
            ans = max(ans,tot+rr[i+k-1][j]-rr[i-1][j]+cc[j+k-1][i]-cc[j-1][i]);
    printf("%d\n", ans);
    return 0;
}