hdu1081 To The Max_最大子矩阵求和问题

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7554    Accepted Submission(s): 3655

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

 

Sample Output

15

 

Source

Greater New York 2001

 

第一道最大子矩阵求和的题目，因为原来没怎么做最大子序列和的题目，搞了一个晚上，囧。。。。。

其实原理和最大子序列和一样，只不过是把二维的数组压缩成一维的来。。。不多说，详见代码：

#include<algorithm>#include<iostream>#include<cstdio>#include<cstring>using namespace std;int sum[105][105];int main(){    int n, cnt, x;    memset(sum, 0, sizeof(sum));    while(scanf("%d", &n) != EOF)    {for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){scanf("%d", &x);sum[i][j] = sum[i - 1][j] + x;//个人觉得这里比较重要，就是把二维压缩成一维的过程，sum[i][j]表示在J列从上向下的数和。}}int Max = -1000000;for (int i = 1; i <= n; ++i){for (int j = i; j <= n; ++j){cnt = 0;for (int k = 1; k <= n; ++k)//表示从第i行到第j行中间和最大的矩阵，这个时候压缩成一维的作用就体现出来了，从第i行到第j行中间的矩阵就变成了一个一维的数组，就转化成了最大子序列和，sum[j][k] - sum[i][k]就表示从k列上，第i行到第j行中间的数的和，接下来的就是最大子序列和的程序。{cnt += sum[j][k] - sum[i][k];if (cnt < 0)cnt = 0;if (cnt > Max)Max = cnt;}}}printf("%d\n", Max);    }    return 0;}