POJ1050(dp)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46788		Accepted: 24774

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

最大子段和的二维版本，把第i行到第j行合并成一行，做法就和一维的一样了，只要枚举i和j，找出最大值即为答案。

 //2016.8.21
 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 const int N = ;
 const int inf = 0x3f3f3f3f;
 int a[N][N];

 int main()
 {
     int n, tmp;
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i < n; i++)
               for(int j = ; j < n; j++)
                   scanf("%d", &a[i][j]);

         int ans = -inf;
         for(int i = ; i < n-; i++)
         {
 //把第j行合并到第i行，求出第i行到第j行的最大子段和**************
             for(int j = i; j < n; j++)
             {
                 tmp = ;
                 for(int k = ; k < n; k++)
                 {
                     if(j > i) a[i][k]+=a[j][k];//把矩阵合并为一维的数组
                     if(tmp > ) tmp += a[i][k];
                     else tmp = a[i][k];
                     ans = max(ans, tmp);
                 }
             }
 //**************************************************************
         }

         cout<<ans<<endl;
     }

     return ;
 }