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  • To the Max_DP

    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 46858   Accepted: 24819

    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

    【题意】 给出一个n*n的方阵,选出一个子矩阵使得里面的值相加最大

    【思路】先不管上下,先看左右,找出值最大的j列到k列,然后确定了以后,就好似成了一列,对这列上下求最大子段和

    dp[i][j][k]=max(dp[i-1][j][k]+tmp,tmp);tmp=第i行j到k列的值之和

    #include<iostream>
    #include<string.h>
    #include<stdio.h>
    using namespace std;
    const int inf=0x7777777;
    const int N=107;
    int a[N][N];
    int dp[N][N][N];
    int main()
    {
        int n;
        while(~scanf("%d",&n))
        {
            memset(dp,0,sizeof(dp));
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    scanf("%d",&a[i][j]);
                }
            }
            int ans=-inf;
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    int tmp=0;
                    for(int k=j; k<=n; k++)
                    {
                        tmp+=a[i][k];
                        dp[i][j][k]=max(dp[i-1][j][k]+tmp,tmp);
                        if(dp[i][j][k]>ans)
                            ans=dp[i][j][k];
                    }
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/iwantstrong/p/5820931.html
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