## poj 1050 To the Max(最大子矩阵权值)

2019年04月13日 算法 ⁄ 共 1943字 ⁄ 字号 评论关闭
To the Max
 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 38220 Accepted: 20161

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

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

8  0 -2```

Sample Output

`15`

Source

```#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
//typedef __int64 ll;
int sum[150][150],dp[150];
int main()
{
int data,n,i,j,k,ans,tp;

memset(sum,0,sizeof sum);
dp[0]=0;
while(~scanf("%d",&n))
{
ans=-INF;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
scanf("%d",&data);
sum[j][i]=sum[j][i-1]+data;
}
}
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
{
for(k=1;k<=n;k++)
{
tp=sum[k][i]-sum[k][i-j-1];
dp[k]=max(tp,dp[k-1]+tp);
ans=max(dp[k],ans);
}
}
}
printf("%d\n",ans);
}
return 0;
}
```