标签:style os io strong for ar amp line
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,直接水过。
#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;
}
ZOJ 1074 To the Max (DP),布布扣,bubuko.com
标签:style os io strong for ar amp line
原文地址:http://blog.csdn.net/u013582254/article/details/38545743