1105 Spiral Matrix

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#pat #蛇形填数

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本文介绍了一种算法，该算法将一组正整数按非递增顺序填充到螺旋矩阵中。文章详细阐述了输入规格、输出规格及样例，并提供了一份参考代码实现。此算法首先对输入序列进行排序，然后根据特定规则确定矩阵尺寸并完成填充。

1105 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 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 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
分析：
将输入序列按从大到小排序，然后填入方阵中，蛇形填数。
参考代码：

#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>

using namespace std;
bool cmp(int a, int b) { return a > b; }
void func(int&m, int&n, int N) {
    int t = sqrt(N);
    for (m = t; m <= N; m++) {
        n = N / m;
        if (m*n == N) { if (n > m) swap(n, m); return; }
    }
}
int main()
{
    freopen("1105.txt", "r", stdin);
    int N, temp, m = 0, n = 0, cnt = 0, r, c;
    vector<int>v;
    scanf_s("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf_s("%d", &temp);
        v.push_back(temp);
    }
    sort(v.begin(), v.end(), cmp);
    func(m, n, N);
    vector<vector<int>>a(m, vector<int>(n, 0));
    a[r=0][c=0] = v[cnt++];
    //蛇形填数的关键处理方法
    while (cnt < N)
    {
        while (c + 1 < n && !a[r][c + 1] && cnt < N) { a[r][c + 1] = v[cnt++]; c++; }
        while (r + 1 < m && !a[r + 1][c] && cnt < N) { a[r + 1][c] = v[cnt++]; r++; }
        while (c - 1 >= 0 && !a[r][c - 1] && cnt < N) { a[r][c - 1] = v[cnt++]; c--; }
        while (r - 1 >= 0 && !a[r - 1][c] && cnt < N) { a[r - 1][c] = v[cnt++]; r--; }
    }

    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (j != 0)cout << " ";
            cout << a[i][j];
        }
        cout << endl;
    }

    return 0;
}