poj1050经典DP的二维形式

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27900		Accepted: 14460

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

题目大意就是给出一个n×n的矩阵，要求在里面找一个子矩阵，使得子矩阵中的元素和是最大的，输出这个最大的元素和

例如对于题目中的矩阵

 0 -2 -7  0

 9  2 -6  2

-4  1 -4  1

-1  8  0 -2

则可以找到子矩阵

 9  2

-4  1

-1  8

其所有元素和为15 是最大的子矩阵

故输出为15

解决这题首先要会求

给n个数，问这n个数的最大子串和是多少？

这一道题就是把上面的问题拓展到了二维，但解决方法类似

枚举所有的行组合，将这些行压缩成一个数列，进行上述操作就可以了

例如对于矩阵

1 2 3 4

2 3 4 1

2 3 1 4

将第一行和第二行压缩结果为3 5 7 5

同理，将第二行第三行压缩结果为4 6 5 5

将一二三行压缩结果为5 8 8 9

•想一想，如果三维呢？三维无非就压缩两次嘛……

刚开始时有个细节没注意到，WA了一次

参考代码：

     1#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[105][105];
int sum[105];
int dp[105];
int main()
{
    int n ;
    while ( cin >> n )
    {
        int i , j , k , t;
        for ( i = 0 ; i < n ; i ++ )
            for ( j = 0 ; j < n ; j ++ )
                cin >> a[i][j] ;
        int sumall = - (1 << 30 );
        for ( i = 0 ; i < n  ; i ++ )
        {
            int sumdp;    
            for ( t = i + 1 ; t <= n ; t ++ )
            {
                for ( j = 0 ; j < n ; j ++ )
                {
                    int sumi = 0 ;
                    for ( k = i ; k < t ; k ++ )
                    {
                        sumi += a[k][j] ;
                    }
                     sum[j] = sumi ;
                     if ( j )
                        sumall = max ( sumall , sumdp = max ( sumdp + sum[j] , sum[j] ) ) ;
                    else
                        sumall = max ( sumall , sumdp = sum[j] ) ;
                }
            }
        }
        cout << sumall << endl ; 
    }
    return 0 ;
}