To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 50466 | Accepted: 26728 |
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
As an example, the maximal sub-rectangle of the array:0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
Input
The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
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4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2 |
Sample Output
1 |
15 |
给出一个矩阵,求出最大子矩阵的和
先求出以(1,1),(x,y)为顶点的每一矩阵和记录为a[x][y],则以(x1,y1),(x2,y2)为顶点矩阵和为a[x2][y2]-a[x1-1][y2]-a[x2][y1-1]+a[x1-1][y1-1]
据此暴力枚举每一子矩阵找出最大值
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#include <cstdio> #include <string> #define MAX(a,b) (a>b)?a:b; int area[102][102]; int main() { int n; scanf("%d", &n ); for (int i = 0; i <= 101; i++) for (int j = 0; j <= 101; j++) area[i][j] = 0; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { scanf("%d", &area[i][j]); area[i][j] = area[i][j] + area[i - 1][j] + area[i][j - 1] - area[i - 1][j - 1]; } int max = -2147483647; for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) for (int k = i; k <= n; k++) for (int l = j; l <= n; l++) max = MAX(max, area[k][l] - area[i - 1][l] - area[k][j - 1] + area[i - 1][j - 1]); printf("%d", max); } |