POJ ---1050 To the Max

时间：2014-03-23 14:39:10 阅读：323 评论：0 收藏：0 [点我收藏+]

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38914		Accepted: 20534

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.

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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int map[105][105], sum[105][105];
 6 int main(){
 7     int n, maxn;
 8     /* freopen("in.c", "r", stdin); */
 9     while(~scanf("%d", &n)){
10         for(int i = 1;i <=n;i ++){
11             for(int j = 1;j <= n;j ++){
12                 scanf("%d", &map[i][j]);
13                 sum[i][j] = sum[i-1][j] + sum[i][j-1] + map[i][j] - sum[i-1][j-1];
14             }
15         }
16         maxn = -100000000;
17         for(int i = 1;i <= n;i ++){
18             for(int j = 1;j <= n;j ++){
19                 for(int k = i+1;k <= n;k ++){
20                     for(int p = j+1;p <= n;p ++)
21                         maxn = max(sum[k][p]-sum[k][j-1]-sum[i-1][p] + sum[i-1][j-1], maxn);
22                 }
23             }
24         }
25         printf("%d\n", maxn);
26     }
27     return 0;
28 }

POJ ---1050 To the Max,布布扣,bubuko.com

POJ ---1050 To the Max

原文：http://www.cnblogs.com/anhuizhiye/p/3618714.html

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