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  • hdu1081 To the max(dp 矩阵压缩)

    To The Max

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8947    Accepted Submission(s): 4323


    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
     
     
    #include<stdio.h>
    #include<string.h>
    #define N 110
    int a[N][N];
    int dp(int b[],int n)
    {
        int sum=0;
        int max=0;
        for(int i=0;i<n;i++)
        {
            sum+=b[i];
            if(sum<0)
            {
                sum=0;
            }
            max=max>sum?max:sum;
        }
        return max;
    }
    int main()
    {
        int n;
        int b[N];
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<n;j++)
                {
                    scanf("%d",&a[i][j]);
                }
            }
            int max=0,sum;
            for(int i=0;i<n;i++)
            {
                memset(b,0,sizeof(b));
                for(int j=i;j<n;j++)
                {
                    for(int k=0;k<n;k++)
                    {
                        b[k]+=a[j][k];
                    }
                    sum=dp(b,n);
                    max=max>sum?max:sum;
                }
            }
            printf("%d
    ",max);
        }    
        return 0;
    }
    //矩阵压缩
    //将i~j行压缩成一行,存在b[]中,然后求 数组b[]的最大值。 
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  • 原文地址:https://www.cnblogs.com/qianyanwanyu--/p/4376982.html
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