【HDU】hdu 1081 （dp_区间dp）-优快云博客

本文介绍了一种使用动态规划解决二维数组中最大子矩形和问题的方法，详细解释了如何将二维数组压缩为一维数组，并求解最大连续子序列和。

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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

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;
    }
}