hdu1081 To The Max （最大子矩阵和）

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4592    Accepted Submission(s): 2165

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

 

分析：通过转换成最大字段和

第一步:s[i][k]=a[1][k]+ …… +a[i][k]

第二步:t[k]=s[j][k]-s[i][k];

第三步:求t[]的最大字段和

 

原因：假设最终的子矩阵和行是i、j；

那么最终是求（a[i][1]+ …… +a[j][1],a[i][2]+ …… +a[j][2], …… ,a[i][n]+ …… +a[j][n]）的最大字段和。

 

#include<iostream>
#include<cstdio>
#define N 110

using namespace std;

int a[N][N],s[N][N];

int max(int A,int B)
{
    return A>B?A:B;
}

int maxsum(int *A,int n)
{
    int i,t,ans;
    ans=A[1];t=0;
    for(i=1;i<=n;i++)
    {
        t=max(A[i],t+A[i]);
        if(ans<t)
            ans=t;
    }
    return ans;
}

int main()
{
    int n,i,j,k,ans,ret,t[N];
    while(scanf("%d",&n)==1)
    {
        for(i=0;i<=n;i++)
            s[i][0]=s[0][i]=a[i][0]=a[0][i]=0;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);
                s[i][j]=s[i-1][j]+a[i][j];
            }
        ans=-100000000;
        for(i=0;i<=n;i++)
        {
            for(j=i+1;j<=n;j++)
            {
                for(k=1;k<=n;k++)
                {
                    t[k]=s[j][k]-s[i][k];
                }
                ret=maxsum(t,n);
                if(ans<ret)
                    ans=ret;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}