POJ 2488 A Knight's Journey(DFS全搜)

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 35837		Accepted: 12216

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

测试数据：
3 7
Scenario #3:
A1B3D2F1G3E2G1F3E1G2E3C2A3B1C3A2C1D3B2D1F2

3 8
Scenario #4:
A1B3C1A2C3D1B2D3E1G2E3C2A3B1D2F1H2F3G1E2G3H1F2H3
题目大意：
人走“日”字型，判断能不能走完全部的方格，若能并输出走的顺序。
思路：就是一个深搜。不过个人认为有个难点：如何记录下所有经过的路径，并DFS返回时该怎么处理。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include<algorithm>
#include<math.h>
using namespace std;
int vis[80][80];
int tx[80],ty[80];
int n,m;
int cx[]= {-1,1,-2,2,-2,2,-1,1};
int cy[]= {-2,-2,-1,-1,1,1,2,2};//此顺次是固定的从下道上，从左到右的次序，并且是对称的
int bj;
void dfs(int s1,int s2,int ans)
{
    tx[ans]=s1;//一定是刚进DFS就存下坐标，否则在return后最后一个路径下标就没法存起来
    ty[ans]=s2;
    if(ans==n*m)//搜索完成的标志
    {
        bj=1;
        return ;
    }
    for(int i=0; i<8; i++)
    {
        int x=cx[i]+s1;
        int y=cy[i]+s2;
        if(x>=1&&y>=1&&x<=n&&y<=m&&!vis[x][y]&&!bj)//注意不要扔掉!bj，如果查找完毕后函数饭回的话，没有此句依然会进行
        {
            vis[x][y]=1;
            dfs(x,y,ans+1);
            vis[x][y]=0;
        }
    }
}
int main()
{
    int cla;
    scanf("%d",&cla);
    for(int gr=1; gr<=cla; gr++)
    {
        bj=0;
        memset(vis,0,sizeof(vis));
        memset(tx,0,sizeof(tx));
        memset(ty,0,sizeof(ty));
        printf("Scenario #%d:\n",gr);
        scanf("%d%d",&n,&m);
        vis[1][1]=1;
        dfs(1,1,1);//便收索边进行步数的计算
        if(bj)
        {
            for(int i=1; i<=n*m; i++)
            {
                printf("%c%d",ty[i]-1+'A',tx[i]);
            }
        }
        else
            printf("impossible");
        printf("\n");
        if(gr!=cla)
            printf("\n");
    }
    return 0;
}