HDU 1081 To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6688    Accepted Submission(s): 3198

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

 

思路：memset(dp,0,sizeof(dp));
memset(temp,0,sizeof(temp));
for(int k = 1;k <= n;k ++)
{
for(int l = i;l <= j;l ++)
temp[k] += map[l][k];
}
dp[1] = temp[1];
for(int k = 1;k <= n;k ++){
if(dp[k - 1] > 0)
dp[k] = dp[k - 1] + temp[k];
else
dp[k] = temp[k];
}
for(int k = 1;k <= n;k ++)
if(dp[k] > max)
max = dp[k];
}

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int map[110][110];
int dp[110];
int temp[110];
int n;
int main()
{
while(~scanf("%d",&n))
{
for(int i = 1;i <= n;i ++)
for(int j = 1;j <= n;j ++)
scanf("%d",&map[i][j]);
int max = 0;
for(int i = 1;i <= n;i ++)
for(int j = i;j <= n;j ++)
{
memset(dp,0,sizeof(dp));
memset(temp,0,sizeof(temp));
for(int k = 1;k <= n;k ++)
{
for(int l = i;l <= j;l ++)
temp[k] += map[l][k];
}
dp[1] = temp[1];
for(int k = 1;k <= n;k ++){
if(dp[k - 1] > 0)
dp[k] = dp[k - 1] + temp[k];
else
dp[k] = temp[k];
}
for(int k = 1;k <= n;k ++)
if(dp[k] > max)
max = dp[k];
}
printf("%d\n",max);
}
return 0;
}

 

 

 

Source

Greater New York 2001