Prefix and the two-dimensional laser bombs

Original title link: https: //www.acwing.com/problem/content/description/101/

A novel laser bombs that can destroy a side length R of all targets within the R square.

There are now on the map N N target, integer the X- i , the Y- i Xi, Yi represents the position of the target on the map, and each has a target value W i the Wi.

Laser bombs served by satellite positioning, but it has a disadvantage that its explosion range, i.e., the side length R sides of the square must be of the R and X , Y X, Y-axis.

If the target is located on the edge of the square blasting, the target will not be destroyed.

Seeking a bomb blew up on the map up to a total value of how much of the target.

Input Format

The first line of the input positive integer N N and R & lt R & lt, represents the number of objects on the side of the square, and the map data separated by spaces.

Next N N rows, each set of input data, each set of data comprises three integers X- I , the Y I , W is I Xi, Yi, the Wi, representing the target X X coordinate, Y Y coordinate and value data separated by a space.

Output Format

Output a positive integer representing up to blow up a bomb on the value of the total number of map objects.

data range

0≤R≤1e9  
0<N≤10000 ,
0≤Xi,Yi≤5000
0≤Wi≤1000

Sample input:

2 1
0 0 1
1 1 1

Sample output:

1
求某点的前缀和：

 

 s[i][j]=g[i-1][j]+g[i][j-1]-g[i-1][j-1]+g[i][j];

二维前缀和：

 

 Red area is all part of the colored area minus the area of ​​green plus yellow area (due to multi-cut)

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=5050;
int n1,r;
int g[maxn][maxn];
int main(){
    cin>>n1>>r;
    int x,y,w;
    int n=r,m=r;
    for(int i=0;i<n1;i++){
        cin>>x>>y>>w;
        x ++ and ++ ;
        n=max(x,n);
        m=max(y,m);
        g[x][y]=w;
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            g[i][j]=g[i-1][j]+g[i][j-1]-g[i-1][j-1]+g[i][j];            
        }
    }
    int ans=0;
    int t;
    for(int i=r;i<=n;i++){
        for(int j=r;j<=m;j++){
            T = G [I] [J] -g [IR] [J] -g [I] [JR] + G [IR] [JR ]; // count is a side is (r-1) is square and, because without the edge
            years = max (year t);
        }
    }
    cout<<ans<<endl;
}