PAT1105:Spiral Matrix

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原文链接：http://www.cnblogs.com/0kk470/p/7911301.html

本文介绍了一种算法，用于将一组正整数按非递增顺序填充到螺旋矩阵中。矩阵的行列数需满足特定条件，并确保按顺时针螺旋方式填充。文章详细解释了输入输出规格、样例及实现步骤。

1105. Spiral Matrix (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 rows and ncolumns, 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 10⁴. 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

思路
题目要求将N个数转换成 m*n 大小的矩阵形式，其中必须满足:
1.m*n = N且满足m-n最小（m >= n）
2.矩阵中的数按从大到小呈顺时针向内螺旋的形式排列，类似一个漩涡一样。

那么有：
1.先将这组数按递减排序。
2.暴力枚举找出满足题目要求1的m和n，构建矩阵二维数组
3.按照顺时针遍历矩阵，将数字一个个输入进去
4.输出。

注意：
1.构建矩阵时可以弄一堵"墙"保证遍历不越界，另外走过的地方也算"墙"(即matrix[i][j] != -1)。
2.用一个数组go[4]表示遍历的每一步（右下左上，顺时针），每当遇到墙（matrix[i][j] != -1,要么是INIT_MAX，要么是之前走过的地方）时改变方向，如此循环。

代码

#include<iostream>
#include<vector>
#include<math.h>
#include<algorithm>
using namespace std;
/*
1.排序
2.找m、n
3.构建矩阵
4.输出
*/
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;
        }
    }
}

转载于:https://www.cnblogs.com/0kk470/p/7911301.html