To The Max
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10184 Accepted Submission(s): 4907
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
/////暴力dp求解0.0
////晚上弄一弄压缩dp玩法0.0
#include<stdio.h>
#include<iostream>
#include<string.h>
int dp[2020][2020];
using namespace std;
int main()
{
int t;
while(~scanf("%d",&t))
{
if(t==0)break;
memset(dp,0,sizeof(dp));
int a;
int output=-100000000;
for(int i=1;i<=t;i++)
{
for(int j=1;j<=t;j++)
{
scanf("%d",&a);
dp[i][j]=a+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
for(int x=1;x<=t;x++)
{
{
for(int y=1;y<=t;y++)
{
if(i>=x&&j>=y)
output=max(output,dp[i][j]+dp[i-x][j-y]-dp[i-x][j]-dp[i][j-y]);
}
}
}
}
}
printf("%d\n",output);
}
}