Solution to poj 1050 “To the Max”

November 14th, 2010 | Tags: ,

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.

To solve this problem, you need to know how to solve the largest sum of successive sub string. Then you could turn this problem from 2-dimension to 1 dimension.

Solution:

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#include <stdio.h>
#include <iostream> 
using namespace std;
 
int getMax(int buf[100],int n)
{
    int temp[101],max=n*(-127);
    memset(temp,0,sizeof(temp));
 
    for(int i=1;i<=n;i++)
    {
        temp[i] = (temp[i-1]>0?temp[i-1]:0)+buf[i];
        if(max<temp[i])
            max=temp[i];
    }
    return max;
}
 
int main(void)
{
    int n,num[101][101],i,j,k,max,temp[101];
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            scanf("%d",&num[i][j]);
    max = -127*n*n;
    for(i=1;i<=n;i++)
    {
        memset(temp,0,sizeof(temp));
        for(j=i;j<=n;j++)
        {
            for(k=1;k<=n;k++)
            {
                temp[k] += num[j][k];
 
            }
            int this_max = getMax(temp,n);
            if(this_max>max)
                max = this_max;
        }
    }
    printf("%d\n",max);
    return 0;
 
}
Leave a comment | Trackback
  1. Hex Bolts
    January 4th, 2011 at 16:19
    Reply | Quote | #1

    没看明白啊！~~~~~~~~

  2. Ethan
    January 5th, 2011 at 12:18
    Reply | Quote | #2

    额，你得先理解最大子串的DP算法，然后再进化一下就回啦～