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[10000][10000];
int N, num[1000000], index = 0;
bool cmp(int a, int b){
    return a > b;
}
int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; i++)
        scanf("%d", &num[i]);
    int sqr = sqrt(N * 1.0);
    int m, n;
    for(n = sqr; N % n != 0; n--);
    m = N / n;
    sort(num, num + N, cmp);
    int k = 0, j = 0, times = 0;
    while(index < N){
        if(N - index == 1){
            G[k][j] = num[index];
            break;
        }
        while(j < n - 1 - times && index < N){
            G[k][j++] = num[index++];
        }
        while(k < m - 1 - 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 = 0; i < m; i++){
        for(int j = 0; j < n; j++){
            if(j == n - 1) 
                printf("%d", G[i][j]);
            else 
                printf("%d ", G[i][j]);
        }
        printf("
");
    }
    cin >> N;
    return 0;
}

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。