hdu 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4825 Accepted Submission(s): 2281

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

Source

Greater New York 2001

//hdu 1003 升级版

http://poj.org/problem?id=1050

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 133
using namespace std;
int dpsum[N][N];
int a[N][N];
int Max,n;
void dp(int &i,int &len)
{
    int r=i+len-1,l=i-1;
    int k,m_ax,b=dpsum[1][r]-dpsum[1][l];
    m_ax=b;
    for(k=2;k<=n;k++)
    {
        if(b<0)
          b=dpsum[k][r]-dpsum[k][l];
        else
         b+=dpsum[k][r]-dpsum[k][l];
        if(b>m_ax)
          m_ax=b;
    }
    Max=max(Max,m_ax);
}
int main()
{
    int i,j,k;
    int l;
    while(scanf("%d",&n)!=EOF)
    {
       for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        {
            scanf("%d",&a[i][j]);
            dpsum[i][j]=dpsum[i][j-1]+a[i][j];
        }
        Max=0;
       for(l=1;l<=n;l++)//求宽度为l的子矩阵和
        {
            j=n-l+1;
            for(i=1;i<=j;i++)
              dp(i,l);
        }
      printf("%d\n",Max);
    }
    return 0;
}