Codeforces Round # 578 (Div. 2) two-dimensional differential template to do

 

 

 

 

 

 Meaning of the title: the n * n matrix, you can select a sub-matrix k * k, and then all the sub-matrix B all into W, ask how you select this sub-matrix so that the resulting matrix in a row W is a full or maximum number of all columns of W

Solution: Consider each row and each column for a particular row, the order to let it all become W, then the upper left corner of the endpoint sub-matrix is ​​in a range, so we can put a value in the range of plus each 1

In order to do two-dimensional differential speed selection, the final answer is the maximum value of the matrix

This question can be used as a two-dimensional differential template

#include<bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)
#define fore(i, s, t) for (int i = s; i < (int)t; i++)
#define fi first
#define se second
#define ll long long

using namespace std;

const int maxn=2e5+5;

const int inf=2e9;


int dif[2005][2005];//difference array

//from (x1,y1) (x2,y2) add val
void add(int x1,int y1,int x2,int y2,int val){
	dif[x1][y1]+=val;
	dif[x1][y2+1]-=val;
	dif[x2+1][y1]-=val;
	dif[x2+1][y2+1]+=val;
} 

string s[maxn];

int main () {
	int n,k;
	cin>>n>>k;
	for(int i=0;i<n;i++){
		cin>>s[i];
	}	
	int res=0;
	for(int i=0;i<n;i++){
		you l = -1, r = -1;
		for(int j=0;j<n;j++){
			if(s[i][j]=='B') {
				if(l==-1) l=j;
				r=j;
			}
		}
		if(l==-1) {
			res++;
			continue;
		}
		if(r-l+1>k) continue;
		add(max(1,i-k+2),max(1,r-k+2),i+1,l+1,1);
	}
	for(int j=0;j<n;j++){
		int l=-1,r=-1;
		for(int i=0;i<n;i++){
			if(s[i][j]=='B') {
				if(l==-1) l=i;
				r=i;
			}
		}
		if(l==-1) {
			res++;
			continue;
		}
		if(r-l+1>k) continue;
		add(max(1,r-k+2),max(j-k+2,1),l+1,j+1,1);
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			dif[i][j]=dif[i][j]+dif[i-1][j]+dif[i][j-1]-dif[i-1][j-1];
			ans=max(ans,dif[i][j]);
		}
	}
	printf("%d\n",ans+res);
}