Input an integer matrix with n rows and m columns, and then input q operations, each operation contains five integers x1, y1, x2, y2, c, where (x1, y1) and (x2, y2) represent a sub-matrix Coordinates of upper left corner and lower right corner.
Each operation adds c to the value of each element in the selected submatrix.
Please output the matrix after all operations.
input format
The first line contains integers n,m,q.
The next n lines, each containing m integers, represent a matrix of integers.
The next q lines, each line contains 5 integers x1, y1, x2, y2, c, representing an operation.
output format
A total of n rows, each row of m integers, represents the final matrix after all operations are completed.
data range
1≤n,m≤1000,
1≤q≤100000,
1≤x1≤x2≤n,
1≤y1≤y2≤m,
−1000≤c≤1000,
−1000≤value of elements in matrix≤1000
Input sample:
3 4 3
1 2 2 1
3 2 2 1
1 1 1 1
1 1 2 2 1
1 3 2 3 2
3 1 3 4 1
Sample output:
2 3 4 1
4 3 4 1
2 2 2 2#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e3 + 10;
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2, int c)
{
b[x1][y1] += c;
b[x2 + 1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
int main()
{
int n, m, q;
cin >> n >> m >> q;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
cin >> a[i][j];
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
insert(i, j, i, j, a[i][j]); //构建差分数组
}
}
while (q--)
{
int x1, y1, x2, y2, c;
cin >> x1 >> y1 >> x2 >> y2 >> c;
insert(x1, y1, x2, y2, c);
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]; //二维前缀和
}
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
printf("%d ", b[i][j]);
}
printf("\n");
}
return 0;
}
If it is extended to two dimensions, we need to add c to the value of each element in the selected sub-matrix of the two-dimensional array, whether it can also achieve the time complexity of O(1). The answer is yes, consider two-dimensional difference.
The a[][] array is the prefix and array of the b[][] array, then b[][] is the differential array of a[][]
Original array: a[i][j]
Let's construct the difference array: b[i][j]
Make a[i][j] in the a array is the sum of the rectangular elements enclosed by the upper left corner (1,1) to the lower right corner (i,j) of the b array.
How to construct the b array?
Let's think backwards.
The purpose of constructing a two-dimensional difference array is to make the operation of adding c to each element in the selected neutron matrix in the original two-dimensional array a, which can be optimized from the time complexity of O(n*n) to O (1)
It is known that the selected sub-matrix in the original array a is a rectangular area surrounded by (x1, y1) as the upper left corner and (x2, y2) as the lower right corner;
Always remember that array a is the prefix and array of array b. For example, the modification of b[i][j] of array b will affect every number in array a from a[i][j] and beyond. .
Assuming that we have constructed the b array, analogous to the one-dimensional difference, we perform the following operations
to add c to the value of each element in the selected sub-matrix
b[x1][y1] + = c;
b[x1,][y2+1] - = c;
b[x2+1][y1] - = c;
b[x2+1][y2+1] + = c;
Performing the above operations on the b array each time is equivalent to:
for(int i=x1;i<=x2;i++)
for(int j=y1;j<=y2;j++)
a[i][j]+=c;