POJ 1050 To the Max
To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 36627 | Accepted: 19262 |
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:
9 2
-4 1
-1 8
and has a sum of 15.
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
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
Sample Output
15
题目大意:最大子矩阵和问题。
解题方法:动态规划。
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; int maze[105][105]; int Subsum[105]; int GetSubsum(int n) { int sum = 0; int Maxsum = -10000000; for (int i = 0; i < n; i++) { if (sum >= 0) { sum += Subsum[i]; } else { sum = Subsum[i]; } Maxsum = max(Maxsum, sum); } return Maxsum; } int main() { int n; scanf("%d", &n); int Maxsum = -10000000; for (int i = 0; i< n; i++) { for (int j = 0; j < n; j++) { scanf("%d", &maze[i][j]); } } for (int i = 0; i < n; i++) { memset(Subsum, 0, sizeof(Subsum)); for (int j = i; j < n; j++) { for (int k = 0; k < n; k++) { Subsum[k] += maze[k][j]; } Maxsum = max(Maxsum, GetSubsum(n)); } } printf("%d\n", Maxsum); return 0; }