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 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

题意：

　　给出一个数组，按照降序，在一个martix中顺时针螺旋排列，要求row >= col 且 abs(row - col)最小。

思路：

　　用一个visited[]数组标记martix是否已经遍历过，然后再按照dirs[]数组进行右下左上，循环填入即可。

Code：

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int i = 0; i < n; ++i) cin >> v[i];
    sort(v.begin(), v.end(), [](int x, int y) -> bool { return x > y; });
    int mid = sqrt(n) + 1;
    int row, col;
    for (int i = mid; i >= 1; --i) {
        if (n % i == 0) {
            row = i;
            col = n / i;
            break;
        }
    }
    if (row < col) {
        int temp = row;
        row = col;
        col = temp;
    }
    vector<vector<int> > matrix(row, vector<int>(col, 0));
    vector<vector<int> > visited = matrix;
    int posX = 0, posY = -1, i = 0;
    int dirs[5] = {0, 1, 0, -1, 0};
    while (i < n) {
        for (int j = 0; j < 4; ++j) {
            while (i < n) {
                int curX = posX + dirs[j];
                int curY = posY + dirs[j + 1];
                if (curX >= 0 && curX < row && curY >= 0 && curY < col &&
                    visited[curX][curY] == 0) {
                    visited[curX][curY] = 1;
                    posX = curX;
                    posY = curY;
                    matrix[posX][posY] = v[i];
                    ++i;
                } else
                    break;
            }
        }
    }
    for (int i = 0; i < row; ++i) {
        cout << matrix[i][0];
        for (int j = 1; j < col; ++j) {
            cout << " " << matrix[i][j];
        }
        cout << endl;
    }
    return 0;
}