108 - Maximum Sum UVA

Maximum Sum

Background

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

The Problem

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size  or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

is in the lower-left-hand corner:

and has the sum of 15.

Input and Output

The input consists of an  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  integers separated by white-space (newlines and spaces). These  integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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].

The output is 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

本题可以利用前缀和解决问题：

对于给定的矩阵grid[m][n]不妨设sum[x][y]代表

以（x，y）为右下角
（0,0）为左上角

的矩形

那么有：sum[x][y] = sum[x-1][y]+sum[x][y-1]-sum[x-1][y-1]+grid[x][y];

sum[0][y]=sum[x][0]=0;

代码如下：

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
int num[110][110];
int sum[110][110];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        int i,j;
        for(i=1;i<=n;++i)
          for(j=1;j<=n;++j)
            scanf("%d",&num[i][j]);
        for(i = 0;i<=n;++i)
        {
            sum[0][i] = 0;
            sum[i][0] = 0;
        }
        for(i = 1;i<=n;++i)
           for(j = 1;j<=n;++j)
             sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + num[i][j];
        int maxn = num[1][1];
        for(i = n;i>=1;--i)
           for(j = n;j>=1;--j)
             for(int k = 0;k<=i;++k)
               for(int l = 0;l<=j;++l)
                  {
                      int temp = sum[i][j] - sum[k][j] - sum[i][l] +sum[k][l];
                      maxn = max(temp,maxn);
                  }
         printf("%d\n",maxn);
    }
    return 0;

}