HDU1081--To The Max

Problem Description

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.

Input

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.

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

 

/*用S[i][j]表示第i行前j个元素的和。则可以按照一维的最大连续和的做法*/#include <iostream>#include <cstdio>using namespace std;#define maxn 108#define inf 0x3f3f3f3fint S[maxn][maxn];//用来存第i行前j个元素的和int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){int key;scanf("%d",&key);S[i][j]=S[i][j-1]+key;}}//接下来就要枚举1列到n列的情况int maxsum=-inf;for(int i=1;i<=n;i++){for(int j=1;j+i<=n+1;j++){int sum=0;for(int k=1;k<=n;k++)//从第一行到最后一行，模拟一维做法即可{sum+=S[k][j+i-1]-S[k][j-1];if(sum>maxsum){maxsum=sum;}if(sum<0)sum=0;}}}printf("%d\n",maxsum);}return 0;}