1050pku To the Max 解题报告
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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.
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
40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2
Sample Output
15
解题思路:这个和一维数组截取一个字段得到的最值是一一样的,只是一维变成二维的,所以要对 数据进行处理,使得像一维数据那样处理,用一个三维的数组来保存一个列里,从i行到j行的和,然 后接着就是像一维数组一样从1~n列,只是这个不是一个数组而已,加了很多数组,这样就可以取他们 中最大的最大即可。(动态规划) 代码如下:
- #include<iostream>
- using namespace std;
- const int Max(101);
- int sum[Max][Max][Max]; //用来保存在一个列里,从i行到j行的和
- int main()
- {
- int n;
- while(cin>>n)
- {
- int data[Max][Max];
- for(int i=1;i<=n;i++)
- {
- for(int j=1;j<=n;j++)
- {
- cin>>data[i][j];
- sum[i][i][j]=data[i][j]; //先初始化为自己那一格的值
- }
- }
- for(int i=1;i<=n;i++)
- {
- for(int j=i+1;j<=n;j++)
- {
- for(int k=1;k<=n;k++)
- {
- sum[i][j][k]=sum[i][j-1][k]+data[j][k]; //计算出在k列从i行到j行的和
- }
- }
- }
- int MaxSum=-9999; //先让初始化他们的最大值
- for(int i=1;i<=n;i++)
- {
- for(int j=i;j<=n;j++)
- {
- int temp=0;
- for(int k=1;k<=n;k++) //和一维数组的得最大的段和是一样的操作
- {
- if(temp>=0)
- temp+=sum[i][j][k];
- else
- temp=sum[i][j][k];
- if(temp>MaxSum)
- MaxSum=temp;
- }
- }
- }
- cout<<MaxSum<<endl;
- }
- return 0;
- }
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