poj1050
To the Max
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 39081 | Accepted: 20639 |
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 /* 2 * File: main.cpp 3 * Author: liaoyu <liaoyu@whu.edu.cn> 4 * 5 * Created on April 1, 2014, 5:34 PM 6 */ 7 8 #include <iostream> 9 #include <stdio.h> 10 #include <stdlib.h> 11 #include <string.h> 12 #include <cmath> 13 #include <algorithm> 14 using namespace std; 15 16 int a[105][105]; 17 18 int maxSubArray(int n, int* a) 19 { 20 int max = 0x80000000; 21 int b = a[0]; 22 for (int i = 1; i < n; i++) { 23 if (b > 0)b += a[i]; 24 else b = a[i]; 25 if (b > max)max = b; 26 } 27 return max; 28 } 29 int b[105]; 30 31 int main() 32 { 33 int n; 34 scanf("%d", &n); 35 int max = 0x80000000; 36 for (int i = 0; i < n; i++) 37 for (int j = 0; j < n; j++) 38 scanf("%d", &a[i][j]); 39 for (int i = 0; i < n; i++) 40 for (int j = i; j < n; j++) { 41 for (int l = 0; l < n; l++) { 42 b[l] = 0; 43 for (int k = i; k <= j; k++) { 44 b[l] += a[l][k]; 45 } 46 } 47 int tmp = maxSubArray(n, b); 48 if (tmp > max)max = tmp; 49 } 50 printf("%d\n", max); 51 }
posted on 2014-04-06 19:17 PolarisSherlock 阅读(118) 评论(0) 编辑 收藏 举报