poj1050

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. 

Input

The input consists of an N * 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

题目意思就是说求最大的子矩阵和；

然后可以通过装换成最大不重叠自序列的模型去做，注意初始化就好了

思路就是先在输入的时候求最大的不重叠的自序列的和，然后再以多行叠加的的思维求最大的不重叠的自序列的和就好了

#include<stdio.h>

int a[105][105];

int main()
{
    int n;
    scanf("%d",&n);
    int sum=-0xfffffff;
    int temp;
    for(int i=1;i<=n;i++){
        temp=0;
        for(int j=1;j<=n;j++){
            scanf("%d",&a[i][j]);
            temp+=a[i][j];
            if(temp>sum)
                sum=temp;
            if(temp<0)
                temp=0;
        }
    }

    for(int i=1;i<n;i++){
        for(int j=i+1;j<=n;j++){
            temp=0;
            for(int k=1;k<=n;k++){
                a[i][k]+=a[j][k];
                temp+=a[i][k];
                if(temp>sum)
                    sum=temp;
                if(temp<0)
                    temp=0;
            }
        }
    }

    printf("%d\n",sum);

    return 0;
}