A1105. Spiral Matrix

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrixis filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and ncolumns, where m and n satisfy the following: m*n must be equal to N; m>=n; and m-n is the minimum of all the possible values.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N. Then the next line contains N positive integers to be filled into the spiral matrix. All the numbers are no more than 10⁴. The numbers in a line are separated by spaces.

Output Specification:

For each test case, output the resulting matrix in m lines, each contains n numbers. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.

Sample Input:

12
37 76 20 98 76 42 53 95 60 81 58 93

Sample Output:

98 95 93
42 37 81
53 20 76
58 60 76

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 int G[][];
 int N, num[], index = ;
 bool cmp(int a, int b){
     return a > b;
 }
 int main(){
     scanf("%d", &N);
     for(int i = ; i < N; i++)
         scanf("%d", &num[i]);
     int sqr = sqrt(N * 1.0);
     int m, n;
     for(n = sqr; N % n != ; n--);
     m = N / n;
     sort(num, num + N, cmp);
     int k = , j = , times = ;
     while(index < N){
         if(N - index == ){
             G[k][j] = num[index];
             break;
         }
         while(j < n -  - times && index < N){
             G[k][j++] = num[index++];
         }
         while(k < m -  - times && index < N){
             G[k++][j] = num[index++];
         }
         while(j > times && index < N){
             G[k][j--] = num[index++];
         }
         while(k > times && index < N){
             G[k--][j] = num[index++];
         }
         k++;
         j++;
         times++;
     }
     for(int i = ; i < m; i++){
         for(int j = ; j < n; j++){
             if(j == n - )
                 printf("%d", G[i][j]);
             else
                 printf("%d ", G[i][j]);
         }
         printf("\n");
     }
     cin >> N;
     return ;
 }

总结：

1、采用旋转填充二维数组的方式的时候要注意，当遇到图形中心是一个时会出现死循环，需要特殊处理一下。如图：

2、测试数据：测试只有1个元素、只有1列元素、边长为奇数的正方形、边长偶数的正方形。

3、超时有可能是因为出现了死循环，而不一定是复杂度不对。

4、方法，设顶点为(x,y)，边长为m、n。填充一圈后变为顶点（x+1, y+1），边长变为m - 2， n - 2。