ZOJ 1074 To the Max （DP)

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Problem

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.

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.

Example

Input

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

Output

15 

水，直接水過。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>using namespace std;int a[110][110];int n,x;int main(){    int sum,maxn;    while(~scanf("%d",&n))    {        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&x);                a[i][j]=a[i-1][j]+x;            }        }        maxn=0;        for(int i=1;i<=n;i++)        {            for(int j=i;j<=n;j++)            {                sum=0;                for(int k=1;k<=n;k++)                {                    int t=a[j][k]-a[i-1][k];                    sum+=t;//                    cout<<"fuck "<<sum<<endl;                    if(sum<0)                        sum=0;                    if(sum>maxn)                        maxn=sum;                }            }        }        printf("%d\n",maxn);    }    return 0;}