1105 Spiral Matrix (25分)(蛇形填数)

1105 Spiral Matrix (25分)

 

This time your job is to fill a sequence of N positive integers into a spiral matrix in non-increasing order. A spiral matrix is filled in from the first element at the upper-left corner, then move in a clockwise spiral. The matrix has m rows and n columns, 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 1. 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


刘汝佳老师的思路，从第一个格子开始走，每次看下一个格子是不是没走过且没越界，
满足条件则走，不满足直接换一个方向。

 1 #include 
 2 #include 
 3 #include 
 4 using namespace std;
 5 
 6 const int maxn = 1e4 + 5, maxe = 100 + 5;
 7 int val[maxn];
 8 int gaze[maxe][maxe];
 9 
10 bool cmp(const int a, const int b) {
11     return a > b;
12 }
13 
14 int main() {
15     int N;
16     scanf("%d", &N);
17     for(int i = 1; i <= N; i ++) {
18         scanf("%d", &val[i]);
19     }
20     sort(val + 1, val + N + 1, cmp);
21     int b = sqrt(N);
22     while(true) {
23         if(N % b == 0) break;
24         b --;
25     }
26     int n = N / b, m = b;
27     int x = 1, y = 0, cnt = 0;
28     while(cnt < N) {
29         while(y + 1 <= m && !gaze[x][y + 1]) gaze[x][++ y] = val[++ cnt];
30         while(x + 1 <= n && !gaze[x + 1][y]) gaze[++ x][y] = val[++ cnt];
31         while(y - 1 > 0 && !gaze[x][y - 1]) gaze[x][-- y] = val[++ cnt];
32         while(x - 1 > 0 && !gaze[x - 1][y]) gaze[-- x][y] = val[++ cnt];
33     }
34     for(int i = 1; i <= n; i ++) {
35         for(int j = 1; j <= m; j ++) {
36             if(j != 1) printf(" ");
37             printf("%d", gaze[i][j]);
38         }
39         printf("\n");
40     }
41     return 0;
42 }

 

第一次做这个题是C语言老师的课堂作业，当时可能写了有七八十行吧，后来在刘汝佳书上看到了一种
精妙的解法，随后忘记了，还有一次在区域赛真题训练赛的时候也写过，当时也是三个人写了好久，绕进去了。
那天打蓝桥杯校赛的时候也写过一个，直接copy的。今天又来了，写了半天发现那个上次被我改出来的点我又忘记了。
真是吐了呀，回顾了一下刘汝佳老师的代码，发现真的是好想而且思路简单。我为什么没想到了，吐了。