hdu 1081 To The Max（二维最大连续和）

To The Max

                                                                  Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                 Total Submission(s): 8735    Accepted Submission(s): 4234

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

 

    题意：

            求出最大子矩阵的和

   题解：

           sum[i][j]表示第i行前j个元素的和，这样第i列到第j列之间的元素和就可以通过sum[k][j]-sum[k][i-1]算出来了，并看成一个元素，这样就成了n个元素的最大连续和（杭电1003题），枚举i，j就可以算出最大值。

代码：

#include 
#include 
#include 
#include 
using namespace std;
int sum[110][110];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int a,ans=-99999999;
       // memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&a);
                sum[i][j]=sum[i][j-1]+a;//第i行前j个元素之和
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                int temp=0;
                for(int k=1;k<=n;k++)//第i列到第j列的情况，最大连续和算法
                {
                    temp+=(sum[k][j]-sum[k][i-1]);
                    if(temp>ans)
                    ans=temp;
                    if(temp<0)
                    temp=0;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}