#1105. Spiral Matrix【模拟】

原题链接

Problem Description：

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: $m\times n$ must be equal to $N$; $m\ge n$; 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 $10^4$. 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

Problem Analysis:

蛇形矩阵经典模型，用偏移量和方向取模来解决顺时针填数问题，首先确定行列大小，并且从大到小排序所有数字，然后循环填数即可。

Code

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 10010;

int n;
int w[N];

int main()
{
    
    
    cin >> n;
    for (int i = 0; i < n; i ++ ) cin >> w[i];
    
    sort(w, w + n, [](int x, int y) {
    
    
        return x > y;
    });
    
    int r, c;
    for (int i = 1; i <= n / i; i ++ )  // 不用求min，i增长，差距越来越小
        if (n % i == 0)
        {
    
    
            r = n / i;
            c = i;
        }
    
    vector<vector<int>> res(r, vector<int>(c));
    
    int dx[] = {
    
    -1, 0, 1, 0}, dy[] = {
    
    0, 1, 0, -1};
    
    // 当前横纵坐标 x, y
    for (int i = 0, x = 0, y = 0, d = 1; i < n; i ++ )
    {
    
    
        res[x][y] = w[i];
        int a = x + dx[d], b = y + dy[d];
        if (a < 0 || a >= r || b < 0 || b >= c || res[a][b])  // 越界或者被占用
        {
    
    
            d = (d + 1) % 4;
            a = x + dx[d], b = y + dy[d];
        }
        x = a, y = b;   
    }
    
    for (int i = 0; i < r; i ++ )
    {
    
    
        cout << res[i][0];
        for (int j = 1; j < c; j ++ )
            cout << ' ' << res[i][j];
        cout << endl;
    }
    return 0;
}