[动态规划]UVA10827 - Maximum sum on a torus

Problem H
Maximum sum on a torus
Input: Standard Input

Output: Standard Output

 

A grid that wraps both horizontally and vertically is called a torus. Given a torus where each cell contains an integer, determine the sub-rectangle with the largest sum. The sum of a sub-rectangle is the sum of all the elements in that rectangle. The grid below shows a torus where the maximum sub-rectangle has been shaded.

 

1	-1	0	0	-4
2	3	-2	-3	2
4	1	-1	5	0
3	-2	1	-3	2
-3	2	4	1	-4

Input

The first line in the input contains the number of test cases (at most 18). Each case starts with an integer N (1≤N≤75) specifying the size of the torus (always square). Then follows N lines describing the torus, each line containing N integers between -100 and 100, inclusive.

 

Output

For each test case, output a line containing a single integer: the maximum sum of a sub-rectangle within the torus.

 

Sample Input                                  Output for Sample Input

2 5 1 -1 0 0 -4 2 3 -2 -3 2 4 1 -1 5 0 3 -2 1 -3 2 -3 2 4 1 -4 3 1 2 3 4 5 6 7 8 9

15 45

Problem setter: Jimmy Mårdell

Special Thanks: Derek Kisman, Md. Kamruzzaman

题意：环形矩阵上的最大子矩阵和。

思路：先复制三个矩阵拼接成一个大的矩阵，然后枚举所求最大子矩阵在第一个矩阵中的左上角，再通过动态规划的方法求出长宽不大于N的最大子矩阵，各种枚举情况中的最大和即为所求解。

#include<iostream>
#include<cstring>
#include<string>

using namespace std;

int arry[1010][1010];

int main()
    {
        string str1,str2;
        while(getline(cin,str1))
            {
                getline(cin,str2);
                int i,j,k;
                memset(arry,0,sizeof(arry));
                int len1=str1.size(),len2=str2.size();
                for(i=1;i<=len1;i++)
                    {
                        for(j=1;j<=len2;j++)
                            {
                                if(str1[i-1]==str2[j-1]) arry[i][j]=arry[i-1][j-1]+1;
                                else arry[i][j]=max(arry[i-1][j],arry[i][j-1]);
                            }
                    }
                cout<<arry[len1][len2]<<endl;
            }
        return 0;
    }