内存:128  时间:1

题目描述

The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

输出

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

样例输入

3
1 1
2 3
4 3

样例输出

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

提示

代码如下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int path[100][2];
int vis[10][10]={0};
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
bool flag=false;
int p,q;
void dfs(int r,int c,int step)
{
    path[step][0]=r;
    path[step][1]=c;
    if(step==p*q)
    {
        flag = true;
        return ;
    }
    for(int i=0;i<8;i++)
    {
        int nx=r+dx[i];
        int ny=c+dy[i];
        if(nx >= 1&&nx<=p&&ny>=1&&ny<=q&&vis[nx][ny]==0&&flag==false)
        {
            vis[nx][ny]=1;
            dfs(nx,ny,step+1);
            vis[nx][ny]=0;
        }
    }
}
int main()
{
    int n,cas=0;
    scanf("%d",&n);
    while(n--)
    {
        flag=false;
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&p,&q);
        vis[1][1]=1;
        dfs(1,1,1);
        printf("Scenario #%d:
",++cas);
        if(flag==true)
        {

            for(int i=1;i<=p*q;i++)
            {
                printf("%c%d",path[i][1]-1+'A',path[i][0]);
            }
        }
        else
            printf("impossible");
        printf("
");
        if(n != 0)
            printf("
");
    }
}

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