POJ 1050 To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 41100 Accepted: 21795

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

Source

Greater New York 2001

#include <iostream>using namespace std;int rect[101][101];int maxvalue;//思想：先压缩到一行，然后横向DP//横向DP 递推公式 maxn[i]=max{0,maxn[i-1]}+a[i];void dp(int n){int line[101];//压缩到一行的数据for(int i=1;i<=n;i++){   for(int j=1;j<=n;j++)   {    //把第i到j行压缩成一行     memset(line,0,sizeof(line));    for(int row = i;row<=j;row++)    {     for(int col = 1;col<=n;col++)     {      line[col]+=rect[row][col];     }    }//end 把第i到j行压缩成一行       //至此，line[1:n]变成了一维的最长子序列    for(int i=2;i<=n;i++)    {     line[i] = (line[i-1]>0?line[i-1]:0)+line[i];     if(line[i]>maxvalue)     {      maxvalue = line[i];     }    }   }//end for j}//end for i}int main(){int n;while(cin>>n){   //数据输入   for(int i=1;i<=n;i++)   {    for(int j=1;j<=n;j++)    {     cin>>rect[i][j];    }   }     maxvalue = 0;   dp(n);     cout<<maxvalue<<endl;}return 0;}