POJ To the Max

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

Source

Greater New York 2001

DP的思想，其实就是暴力枚举上下界，然后把二维问题化为一维的来做

#include<stdio.h>
#include<string.h>

int min(int a,int b)
{
  return a<b?a:b;
}

int max(int a,int b)
{
  return a>b?a:b;
}

int array[105];
int sum[105][105];
int num[105][105];

int main()
{
  int n;
  while(~scanf("%d",&n))
  {
    sum[0][0]=0;
    array[0]=0;
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        scanf("%d",&num[i][j]);
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        sum[i][j]=sum[i-1][j]+num[i][j];
    int ans=-0x3f3f3f3f;
    for(int up=1;up<=n;up++)
      for(int down=up;down<=n;down++)
      {
        for(int i=1;i<=n;i++)
          array[i]=array[i-1]+(sum[down][i]-sum[up-1][i]);
        int mins=0;//保存1-(i-1)里面的最小值
        for(int i=1;i<=n;i++)
        {
            ans=max(ans,array[i]-mins);
          mins=min(mins,array[i]);
        }
      }
    printf("%d\n",ans );
  }
  return 0;
}