ZOJ 1074 To the Max（DP 最大子矩阵和）

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：

 1 # include<stdio.h>
 2 # include<string.h>
 3 # define N 101
 4 int main(){
 5     int a[N][N],b[N];
 6     int n,i,j,k;
 7     while(scanf("%d",&n)!=EOF)
 8     {
 9         for(i=0;i<n;i++)
10             for(j=0;j<n;j++)
11                 scanf("%d",&a[i][j]);
12             int max= -12345;
13             for(i=0;i<n;i++)    //第i行到第j行的最大子矩阵和
14             {
15                 memset(b,0,sizeof(b));
16                 for(j=i;j<n;j++)
17                 {
18                     int sum=0;
19                     for(k=0;k<n;k++)
20                     {
21                         b[k] += a[j][k];
22                         sum += b[k];
23                         if(sum<0)  sum=b[k];
24                         if(sum>max)  max=sum;
25                     }
26                 }
27             }
28             printf("%d\n",max);
29     }
30     return 0;
31 }