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): 7620    Accepted Submission(s): 3692

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

 

 

　　这道题目是求二维数组的最大子矩阵的和，最大子矩阵一定是在1~n行之间，所以要任选连续的几行压缩成一位数组求最大连续子段和。

代码:

#include <iostream>
#include <cstdio>
using namespace std;
#define N 105
int arr[N][N],b[N];
int dp(int *a,int m)    //求一维数组的最大子段和
{
    int i,sum,max;
    sum = 0;
    max = 0;
    for(i=0; i<N; i++)
    {
        sum += a[i];
        if(sum<0)
            sum = 0;
        if(sum>max)
            max = sum;
    }
    return max;
}
int main()
{
    int i,j,k,n,sum,max;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                scanf("%d",&arr[i][j]);
        max = 0;
        for(i=0; i<n; i++)
        {
            memset(b,0,sizeof(b));
            for(j=i; j<n; j++)
            {
                for(k=0; k<n; k++)
                    b[k] += arr[j][k];
                sum = dp(b,n);
                if(sum>max)
                    max = sum;
            }
        }
        printf("%d
",max);
    }
    return 0;
}