ZOJ 1074 To the Max（DP 最大子矩阵和）

To the Max

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Problem

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.

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.

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 

代码如下：

# include<stdio.h>
# include<string.h>
# define N 101
int main(){
    int a[N][N],b[N];
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                scanf("%d",&a[i][j]);
            int max= -12345;
            for(i=0;i<n;i++)    //第i行到第j行的最大子矩阵和
            {
                memset(b,0,sizeof(b));
                for(j=i;j<n;j++)
                {
                    int sum=0;
                    for(k=0;k<n;k++)
                    {
                        b[k] += a[j][k];
                        sum += b[k];
                        if(sum<0)  sum=b[k];
                        if(sum>max)  max=sum;
                    }
                }
            }
            printf("%d\n",max);
    }
    return 0;
}