POJ 2488 DFS 模拟马的跳动

A Knight's Journe

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40970		Accepted: 13938

Description

Background
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?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

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, . . .

Output

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.

Sample Input

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

注意马的跳动列数是按照字典序调的，也就是说必须按照A B C.....A B....这样

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 30;
int vis[maxn][maxn];
int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int n,m,flag;

struct Path{
     int x,y;
}p[maxn];

void dfs(int x,int y,int step){
     if(flag)
        return;

        p[step].x=x;
        p[step].y=y;

        if(step==n*m)
        {
            flag = 1;
            return ;
        }

        Path next;
        for(int i = 0;i<8;i++)
        {
            next.x = x+dir[i][0];
            next.y = y+dir[i][1];

            if(next.x >=1&&next.x<=m&&next.y>=1&&next.y<=n&&!vis[next.x][next.y])
            {
                vis[next.x][next.y]=1;
                dfs(next.x,next.y,step+1);
                vis[next.x][next.y]=0;
            }
           }
            return ;
     }
int main()
{
    int T;
    cin>>T;
    for(int t = 1;t<=T;t++){
        cin>>n>>m; //输入行数和列数，行数是y的边界，列数是x的边界
        memset(vis,0,sizeof(vis));
        memset(p,0,sizeof(p));
        flag=0;
        vis[1][1]=1;
        dfs(1,1,1);
       printf("Scenario #%d:
", t);
       if(flag){
        for(int i = 1;i<=n*m;i++)
            printf("%c%d",p[i].x-1+'A',p[i].y);
            cout<<endl;
       }
       else cout<<"impossible"<<endl;
       if(t!=T) cout<<endl;
    }
    return 0;
}

注意模拟的是二维数组的坐标，不要和数学坐标弄淆。

POJ 2488 DFS 模拟 马的跳动

POJ 2488 DFS 模拟马的跳动