【乱搞】【AOJ-298】Rings of square grid

Description

Any square grid can be viewed as one or more rings, one inside the other. For example, as shown in figure (a), a 5 X 5 grid is made of three rings, numbered 1,2 and 3 (from outside to inside.) A square grid of size N is said to be sorted, if it includes the values from 1 to N2 in a row-major order, as shown in figure (b) for N = 4. We would like to determine if a given square grid can be sorted by only rotating its rings. For example, the grid in figure (c) can be sorted by rotating the first ring two places counter-clockwise(逆时针), and rotating the second ring one place in the clockwise direction.

Input

Your program will be tested on one or more test cases. The first input line of a test case is an integer N which is the size of the grid. N input lines will follow, each line made of N integer values specifying the values in the grid in a row-major order. Note than 0 < N ≤ 1, 000 and grid values are natural numbers less than or equal to 1,000,000.
The end of the test cases is identified with a dummy test case with N = 0.

Output

For each test case, output the result on a single line using the following format:
k.result
Where k is the test case number (starting at 1,)  is a single space, and result
is "YES" or "NO" (without the double quotes.)

Sample Input

4
9 5 1 2
13 7 11 3
14 6 10 4
15 16 12 8
3
1 2 3
5 6 7
8 9 4
0

 

Sample Output

1. YES
2. NO

 

思路：

记录需要比较的元素与现有元素进行比较 错误就退出

 

参考代码

#include<stdio.h>
#define LEN1 1000+10
#define LEN2 20000
long int m1[LEN1][LEN1],m2[LEN1][LEN1];
long int a[LEN2],b[LEN2];
int main()
{

    int n,m=1;
    int i,j,k;
    int c,flag,q;
    int s;
    while(scanf("%d",&n)&&n)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=n; j++)
            {
                m1[i][j]=(i-1)*n+j;
                scanf("%ld",&m2[i][j]);
            }

            c=(n+1)/2;
            flag=0;
            for(k=1;k<=c;k++)
            {
                q=0;
                for(i=k;i<n+1-k;i++)
                {
                    a[q]=m1[k][i];
                    b[q]=m2[k][i];
                    q++;
                }
                for(i=k;i<=n+1-k;i++)
                {
                    a[q]=m1[i][n+1-k];
                    b[q]=m2[i][n+1-k];
                    q++;
                }
                for(i=n-k;i>=k;i--)
                {
                    a[q]=m1[n+1-k][i];
                    b[q]=m2[n+1-k][i];
                    q++;
                }
                for(i=n-k;i>k;i--)
                {
                    a[q]=m1[i][k];
                    b[q]=m2[i][k];
                    q++;
                }

                s=0;
                for(i=0;i<q;i++)
                {
                    if(a[i]==b[0])
                    {
                        s=i;
                        break;
                    }
                }
                j=0;
                for(i=s;i<q;i++)
                    if(a[i]!=b[j++])
                    {
                        flag=1;
                        break;
                    }
                    if(!flag)
                    {
                        for(int i=0;i<s;i++)
                            if(a[i]!=b[j++])
                            {
                                flag=1;
                                break;
                            }
                    }

            }
            if(!flag)
                printf("%d. YES
",m++);
            else
                printf("%d. NO
",m++);
    }
    return 0;
}