POJ1050 To the Max

时间：2016-07-29 20:58:34 阅读：279 评论：0 收藏：0 [点我收藏+]

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

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。

 1 /**/
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 using namespace std;
 8 const int mxn=125;
 9 int a[mxn][mxn];
10 int n;
11 int sum;
12 int ans=0;
13 int main(){
14     int i,j;
15     scanf("%d",&n);
16     for(i=1;i<=n;i++)
17       for(j=1;j<=n;j++){
18           scanf("%d",&a[i][j]);
19           a[i][j]+=a[i][j-1];
20       }
21     for(i=0;i<n;i++)
22      for(j=i;j<=n;j++){
23          sum=0;
24          for(int k=0;k<=n;k++){
25              sum+=a[k][j]-a[k][i];
26              if(sum<0)sum=0;
27              ans=max(ans,sum);
28          }
29      }
30     printf("%d\n",ans);
31 }

POJ1050 To the Max

原文：http://www.cnblogs.com/SilverNebula/p/5719439.html

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