[南阳OJ-No.33]蛇形填数|在n*n方陈里填入1,2,...,n*n,要求填成蛇形。

南阳OJ-No.33

时间限制3000ms,空间限制65535KB,难度3

描述

在n*n方陈里填入1,2,…,n*n,要求填成蛇形。例如n=4时方阵为：

10 11 12 1
9 16 13 2
8 15 14 3
7 6 5 4

输入

直接输入方陈的维数，即n的值。(n<=100)

输出

输出结果是蛇形方陈。

样例输入

3

样例输出

7 8 1
6 9 2
5 4 3

---

思路：4 连通问题
n=1，特殊情况
n>2，
我先画出了n=2,3,4,5,6的蛇形数组进行观察，数组构建过程是从右上角的数字1开始，然后按照顺时针方向依次构建。方阵中每个格子相邻的都有四个格子。观察构建顺序，数字存放顺序为顺时针方向，即下、左、上、右，进行循环判断填充，注意判断数组下标不越界

---

JAVA

时间83，内存1475

import java.util.Scanner;

public class Main {
    public static Scanner cin = new Scanner(System.in);
    public static int N = cin.nextInt(), x = 0, y = N-1, count = 1;
    public static int[][] num = new int[N][N];

    static {
        for (int x=0; xfor (int y=0; y0;
            } 
        }

        num[x][y] = 1;
    }

    public static void print(int[][] num) {
        for (int x=0; xfor (int y=0; yout.print(num[x][y] + " ");
            }
            System.out.println();
        }
    }

    public static void main(String[] args) throws Exception {   
        while(count < N*N) {
            while(x1 && num[x+1][y]==0)
                num[++x][y] = ++count;

            while(y>0 && num[x][y-1]==0)
                num[x][--y] = ++count;

            while(x>0 && num[x-1][y]==0)
                num[--x][y] = ++count;

            while(y1 && num[x][y+1]==0)
                num[x][++y] = ++count;
        }
        print(num);
    }
}

这个是在南阳OJ中扒出来的以为大神，不懂啊(⊙﹏⊙)
时间36，内存676

import java.util.Scanner;

public class Main{
    public static void main(String[] args) {
        int i,j,h = 0;
        short num = 1;
        Scanner scanner = new Scanner(System.in);
        short n = scanner.nextShort();
        short[][] array = new short[n][n];
        int hmax = (n+1)>>1;
        for ( h = 0; h <= hmax; h++) {
            if (h==hmax&&n%2!=0) {
                array[n>>1][n>>1] = num;
            }else {
                for ( i =h; i < n-1-h; i++) {
                    array[i][n-1-h]=num++;
                }
                for ( i = n-1-h; i >h ; i--) {
                    array[n-1-h][i]=num++;
                }
                for ( i = n-1-h; i >h ; i--) {
                    array[i][h]=num++;
                }
                for ( i = h; i < n-1-h; i++) {
                    array[h][i] = num++;
                }
            }
        }
        StringBuffer sb = new StringBuffer();
        for ( i = 0; i < array.length; i++) {
            for ( j = 0; j < array[i].length; j++) {
                sb.append(array[i][j]).append(" ");
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
}        

C++

时间4，内存240

#include 
using namespace std;

int main()
{
  int N, x, y, count;

  cin >> N;
  x = 0;
  y = N-1;
  count = 1;

  int num[N][N];

  for (int x=0; xfor (int y=0; y0;
        } 
  }

  num[x][y] = 1;

  while(count < N*N) {
        while(x1 && num[x+1][y]==0)
            num[++x][y] = ++count;

        while(y>0 && num[x][y-1]==0)
            num[x][--y] = ++count;

        while(x>0 && num[x-1][y]==0)
            num[--x][y] = ++count;

        while(y1 && num[x][y+1]==0)
            num[x][++y] = ++count;
  }

  for (int x=0; xfor (int y=0; y" ";
        }
        cout << endl;
  }

  return 0;
}