【HDU】hdu 1081 （dp_区间dp）

 

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

 

 

 

这是一个动态规划问题，onedim[ ]表示将i 到 j 行的数字压缩到一行，然后就可以将问题转换到求最大子序列的和的问题，进一步求出问题的解，特别要注意数组的初始化问题，每次讲二维数组压缩到一维数组的时候，都要将数组初始化为0；然后进行压缩，每次压缩之后就可以取最大值，进而求出答案

代码：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>

using namespace std;
int arrsize;
int arrnumber[105][105];
int onedim[105];
void trans(int srow ,int enrow )///将二维矩阵转化为一维的；然后求最大连续子序列的和
{
    memset(onedim,0,sizeof(onedim));
    for ( int j = 0; j < arrsize; j++ )
    {
        for ( int i = srow; i<= enrow; i++ )
        {
            onedim[j] += arrnumber[i][j];
        }
    }
}
int maxsum ()///求最大连续子序列的和
{
    int bigsum[105];
    memset(bigsum,0,sizeof(bigsum));
    bigsum[0] = onedim[0];
    for ( int i = 1; i < arrsize; i++ )
    {
        if ( bigsum[i-1] > 0) bigsum[i]=bigsum[i-1]+onedim[i];
        else bigsum[i] = onedim[i];
    }
    int ans = 0;
    for ( int i = 1; i < arrsize; i++ )
    {
        ans = max(ans,bigsum[i]);
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while ( cin >> arrsize )
    {
        for ( int i = 0; i < arrsize; i++ )
        {
            for ( int j = 0; j < arrsize; j++ )
            {
                cin >> arrnumber[i][j];
            }
        }
        int ans = 0;
        for ( int i = 0 ;i < arrsize; i++ )
        {
            for ( int j = i; j < arrsize; j++ )
            {
                trans(i,j);
                ans = max(ans,maxsum());
            }
        }
        cout << ans << endl;
    }
}