POJ1050 To the Max

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46562		Accepted: 24627

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

Greater New York 2001

 

二维的求连续序列和。和一维思路基本一样。枚举左右列数边界，然后从上到下逐行累计限制范围内的和，小于0就重置。

/**/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int mxn=125;
int a[mxn][mxn];
int n;
int sum;
int ans=0;
int main(){
    int i,j;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
      for(j=1;j<=n;j++){
          scanf("%d",&a[i][j]);
          a[i][j]+=a[i][j-1];
      }
    for(i=0;i<n;i++)
     for(j=i;j<=n;j++){
         sum=0;
         for(int k=0;k<=n;k++){
             sum+=a[k][j]-a[k][i];
             if(sum<0)sum=0;
             ans=max(ans,sum);
         }
     }
    printf("%d\n",ans);
}