1105. Spiral Matrix (25)

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

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 10⁴

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.

12
37 76 20 98 76 42 53 95 60 81 58 93

98 95 93
42 37 81
53 20 76
58 60 76

思路
題目要求將N個數轉換成 m*n 大小的矩陣形式，其中必須滿足:
1.m*n = N且滿足m-n最小（m >= n）
2.矩陣中的數按從大到小呈順時針向內螺旋的形式排列，類似一個漩渦一樣。

那麽有：
1.先將這組數按遞減排序。
2.暴力枚舉找出滿足題目要求1的m和n，構建矩陣二維數組
3.按照順時針遍歷矩陣，將數字一個個輸入進去
4.輸出。

註意：
1.構建矩陣時可以弄一堵"墻"保證遍歷不越界，另外走過的地方也算"墻"(即matrix[i][j] != -1)。
2.用一個數組go[4]表示遍歷的每一步（右下左上，順時針），每當遇到墻（matrix[i][j] != -1,要麽是INIT_MAX，要麽是之前走過的地方）時改變方向，如此循環。

代碼

#include<iostream>
#include<vector>
#include<math.h>
#include<algorithm>
using namespace std;
/*
1.排序
2.找m、n
3.構建矩陣
4.輸出
*/
vector<vector<int>> go ={{0,1},{1,0},{0,-1},{-1,0}};//右下左上
const int INIT_MAX = pow(2,30);
bool cmp(const int a,const int b)
{
    return a > b;
}

int main()
{
    int N;
    while(cin >> N)
    {
        vector<int> num(N);
        for(int i = 0;i < N;i++)
        {
            cin >> num[i];
        }
        sort(num.begin(),num.end(),cmp);

        //find min(m - n)
        int m,n,curmin = INIT_MAX;
        for(int i = N;i >= sqrt(N);i--)
        {
            if(i * (N/i) == N && i - (N/i) < curmin)
            {
                 m = i;
                 n = N/i;
                 curmin = m - n;
            }
        }
        //build matrix
        vector<vector<int>> matrix(m + 2,vector<int>(n + 2,-1));
        for(int i = 0;i <= n + 1;i++)
        {
            matrix[0][i] = matrix[m + 1][i] = INIT_MAX;
        }
        for(int i = 0;i <= m + 1;i++)
        {
            matrix[i][0] = matrix[i][n + 1] = INIT_MAX;
        }
        int a = 1,b = 1,dir = 0;
        matrix[a][b] = num[0];
        for(int i = 1;i < num.size();i++)
        {
           if(matrix[a+go[dir][0]][b+go[dir][1]] != -1)
           {
              dir++;
              if(dir > 3)
                dir = 0;
           }
           a += go[dir][0];
           b += go[dir][1];
           matrix[a][b] = num[i];
        }
        //output
        for(int i = 1;i <= m;i++)
        {
            for(int j = 1;j <= n;j++)
            {
                if(j != 1)
                  cout << " ";
                cout << matrix[i][j];
            }
            cout << endl;
        }
    }
}

PAT1105:Spiral Matrix