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。

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const int maxn = 1e4 + 5, maxe = 100 + 5;
int val[maxn];
int gaze[maxe][maxe];

bool cmp(const int a, const int b) {
    return a > b;
}

int main() {
    int N;
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &val[i]);
    }
    sort(val + 1, val + N + 1, cmp);
    int b = sqrt(N);
    while(true) {
        if(N % b == 0) break;
        b --;
    }
    int n = N / b, m = b;
    int x = 1, y = 0, cnt = 0;
    while(cnt < N) {
        while(y + 1 <= m && !gaze[x][y + 1]) gaze[x][++ y] = val[++ cnt];
        while(x + 1 <= n && !gaze[x + 1][y]) gaze[++ x][y] = val[++ cnt];
        while(y - 1 > 0 && !gaze[x][y - 1]) gaze[x][-- y] = val[++ cnt];
        while(x - 1 > 0 && !gaze[x - 1][y]) gaze[-- x][y] = val[++ cnt];
    }
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= m; j ++) {
            if(j != 1) printf(" ");
            printf("%d", gaze[i][j]);
        }
        printf("\n");
    }
    return 0;
}

 

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。