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): 10747    Accepted Submission(s): 5149

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

 

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迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int fun(int b[110],int n)
{
    int i,sum=0,max=-130;
    for (i = 1; i <= n; i++)
    {
        if (sum > 0)sum += b[i];
        else sum = b[i];
        if (max < sum)max = sum;
    }
    return max;
}
int main()
{
    int i,j,k,sum,max,n,a[110][110], b[110];
    while (cin>>n)
    {
        sum = 0,max = -130;
        for (i = 1; i <= n; i++)
            for (j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
        for (i = 1; i <= n; i++)
        {
            memset(b,0,sizeof(b));
            for (j = i; j <= n; j++)
            {
                for (k = 1; k <= n; k++)
                    b[k]+=a[j][k];
                sum = fun(b,n);
                if (max < sum)max = sum;
            }
        }
        printf("%d
", max);
    }
    return 0;
}