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 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 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

#include<iostream>
#include<vector>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iomanip>
using namespace std;
int cmp(int a, int b)
{
  return a > b;
}
/*public class Helix {

static int N = 0;

public static int Helix(int x, int y) {
if ((x==1) && (y==1)) return 1;
if ((x-y>=1)&&(x+y<=N)) return (1 + Helix(x+1,y));
if ((x-y>=1)&&(x+y>N)) return (1 + Helix(x, y+1)); 
if ((x-y<1)&&(x+y<=N+1)) return (1 + Helix(x, y-1));
if ((x-y<1)&&(x+y)>N+1) return (1 + Helix(x-1,y));
else return 0;} 

public static void main(String[] args) {

N = //输入int值;
for (int i= 1; i <= N; i++){
for (int j= 1; j <= N; j++)
System.out.print(Helix(i,j) + "\t");
System.out.println(""); }


}}这个就是先把矩阵每个位置二元组转换成一维时对应的数值求出来，我转别人的，不是自己想的，实质就是迷宫求解问题
*/
int main()
{
  int N;
  cin >> N;
  int t;
  int N1 = N;
  vector<int> v;
  while (N--)
  {
    cin >> t;
    v.push_back(t);
  }
  sort(v.begin(), v.end(), cmp);//非递增
  int i;
  for (i = sqrt(N1); i <= N1; i++)
  {
    if (N1%i == 0)
      break;
  }//m和n肯定一个小于等于N平方根另一个大于等于N平方根
  int m = max(i,N1/i), n = N1 / m;
  int**a = new int*[m];
  for (int i = 0; i < m; i++)
  {
      a[i] = new int[n];
  }
  for (int i = 0; i < m; i++)
  {
    for (int j = 0; j < n; j++)
    {
      a[i][j] = 0;
    }
  }//必须初始化为0
  
  i = 0;
  int x = 0, y = 0;
  while (i < N1)
  {
    while (y < n&&!a[x][y])
      a[x][y++] = v[i++];
    y--;
    x++;
    while (x < m && !a[x][y])
      a[x++][y] = v[i++];
    x--;
    y--;
    while (y >= 0 && !a[x][y])
      a[x][y--] = v[i++];
    y++;
    x--;
    while (x >= 0 && !a[x][y])
      a[x--][y] = v[i++];
    x++;
    y++;
  }//螺旋放置的过程：先声明横坐标x和纵坐标y，然后类似走迷宫的方式，若前面能走（即判断下一个位置是否为0，不为0就不能进行赋值了)
  //分为四个过程，右下左上，依次进行即可
  for (int i = 0; i < m; i++)
  {
    for (int j = 0; j < n; j++)
    {
      cout << a[i][j];
      if (j != n - 1)
        cout << " ";
      else
        cout << endl;
    }
    
  }
  return 0;
}