AcWing 796. The sum of submatrices [two-dimensional prefix and c++ detailed solution]

Two-dimensional prefix sum

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topic

Enter an integer matrix with n rows and m columns, and then enter q queries. Each query contains four integers x1, y1, x2, y2, which represent the coordinates of the upper left corner and the lower right corner of a sub-matrix.

For each query, output the sum of all numbers in the sub-matrix.

Input format The
first line contains three integers n, m, q.

The next n rows, each row contains m integers, representing a matrix of integers.

Next line q, each line contains four integers x1, y1, x2, y2, representing a set of queries.

Output format There are
q lines, and each line outputs a query result.

Data range
1≤n, m≤1000,
1≤q≤200000,
1≤x1≤x2≤n,
1≤y1≤y2≤m,
−1000≤The value of the element in the matrix≤1000
Input example:
3 4 3
1 7 2 4
3 6 2 8
2 1 2 3
1 1 2 2
2 1 3 4
1 3 3 4
Output example:
17
27
21

Two-dimensional prefix and derivation

As shown:

The purple area refers (1,1)to (i,j-1)the rectangular area from the upper left corner to the lower right corner, and the green area refers (1,1)to (i-1, j )the rectangular area from the upper left corner to the lower right corner. The area of ​​each color rectangle represents the sum of its surrounding elements.

From the figure, we can easily see that the area of ​​the entire outer blue rectangle s[i][j]= green area s[i-1][j]+ purple area s[i][j-1]-repeated red area s[i-1][j-1]+ small square area a[i][j];

Therefore, the two-dimensional prefix and preprocessing formula

s[i] [j] = s[i-1][j] + s[i][j-1 ] + a[i] [j] - s[i-1][ j-1]

Next regression to seek to (x1,y1)the upper left corner and to the (x2,y2)elements of the matrix and the lower right corner.

As shown:

Purple area refers to the ( 1,1 )upper left corner of the (x1-1,y2)rectangular area in the lower right corner, the yellow area refers to the (1,1)upper left corner of the (x2,y1-1)rectangular area of the lower right corner;

Not difficult to launch:

The area of ​​the green rectangle = the entire outer area s[x2, y2]-the yellow area s[x2, y1 - 1]-the purple area s[x1 - 1, y2]+ the red area repeatedly subtracteds[x1 - 1, y1 - 1]

So the conclusion of the two-dimensional prefix sum is:

To (x1, y1)the upper left corner, (x2, y2)for the child and for the lower right corner of the matrix:
s[x2, y2] - s[x1 - 1, y2] - s[x2, y1 - 1] + s[x1 - 1, y1 - 1]

to sum up:

Summary of personal experience of prefix sum and difference

Code:

#include<iostream>
#include<cstdio>
using namespace std;
const int N=1010;
int n,m,q;
int a[N][N],s[N][N];
int main()
{
    
    
    scanf("%d %d %d",&n,&m,&q);
    for(int i=1;i<=n;i++)
    {
    
    
        for(int j=1;j<=m;j++)
        {
    
    
            scanf("%d",&a[i][j]);
        }
    }
    for(int i=1;i<=n;i++)
    {
    
    
        for(int j=1;j<=m;j++)
        {
    
    
            s[i][j]=s[i-1][j]+s[i][j-1]+a[i][j]-s[i-1][j-1];  //求前缀和
        }
    }
    while(q--)
    {
    
    
        int x1,y1,x2,y2;
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        printf("%d\n",s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1]);
    }
    return 0;
}