To the Max - POJ 1050 dp

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 39758 Accepted: 21010

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

题意：找到和最大的矩阵并输出最大的和。

思路：我们先看一下如果是一维的怎么办，sum[i]=  sum[i-1]<=0 ?  num[i] ：sum[i-1]+num[i]。

           这道题只要把当做是n*(n-1)/2个一维的数，然后去算就可以了，复杂度应该是小于o(n^3)，但是最后时间出来是16ms，看来数据比较水。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;int num[110];int res1[110][110];int res2[110][110];int main(){ int n,i,j,k,ans=-100000000;  scanf("%d",&n);  for(k=1;k<=n;k++)  { for(i=1;i<=n;i++)     scanf("%d",&num[i]);    for(i=1;i<=n;i++)     for(j=i;j<=n;j++)     { res1[i][j]=res1[i][j-1]+num[j];       if(res2[i][j]<=0)        res2[i][j]=res1[i][j];       else        res2[i][j]=res2[i][j]+res1[i][j];       if(res2[i][j]>ans)        ans=res2[i][j];     }  }  printf("%d\n",ans);}