HDU 1081：To The Max

HDU 1081：To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10747    Accepted Submission(s): 5149

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

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​​

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迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

#include#include#includeusing namespace std;int fun(int b[110],int n){ int i,sum=0,max=-130; for (i = 1; i <= n; i++) { if (sum > 0)sum += b[i]; else sum = b[i]; if (max < sum)max = sum; } return max;}int main(){ int i,j,k,sum,max,n,a[110][110], b[110]; while (cin>>n) { sum = 0,max = -130; for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) scanf("%d", &a[i][j]); for (i = 1; i <= n; i++) { memset(b,0,sizeof(b)); for (j = i; j <= n; j++) { for (k = 1; k <= n; k++) b[k]+=a[j][k]; sum = fun(b,n); if (max < sum)max = sum; } } printf("%d\n", max); } return 0;}

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