hdu 1081 To The Max (动态规划)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5917    Accepted Submission(s): 2807


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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].

 

[align=left]Output[/align]
Output the sum of the maximal sub-rectangle.

 

[align=left]Sample Input[/align]

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1 8 0 -2

 

[align=left]Sample Output[/align]

15

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 110
int f

;
int main(){
int n,i,j,k,tmp;
int ans;
while(cin>>n){
memset(f,0,sizeof(f));
for(i=1;i<=n;i++)
for(j=1;j<=n;j++){
cin>>k;
f[i][j]=f[i][j-1]+k;//求和这一列1-j的和
}
ans=-99999999;
for(i=1;i<=n;i++)//枚举1-n列
for(j=1;j<=i;j++){
tmp=-1;
for(k=1;k<=n;k++){//枚举1-n行
if(tmp>0)tmp+=f[k][i]-f[k][j-1];//上一个结果大于0就累加进去
else tmp=f[k][i]-f[k][j-1];//矩形（k,j-1)-(k,j)的和
ans=ans>tmp?ans:tmp;
}
}
cout<<ans<<endl;
}
return 0;
}