poj2488(dfs+按字典序输出路径)

 题目链接：http://poj.org/problem?id=2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 52426		Accepted: 17787

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

输出路径的寻路问题

字典序只需要调整方向数组的优先级即可

p行q列，但输出时先输出q所对的坐标，就像给你(x,y),让你输出(y,x),算是小坑。。。（也可能是我太蠢了QAQ）

因为字典序最小，所以第一个一定是A1，即从A1开始

因为已经调整了方向数组的优先级，所以找到的第一个路径就是答案

代码附上：

#include<cstdio>
#include<iostream>
#include<queue>
#include<string>
#include<cstring>
#define sca(a) scanf("%d",&a );
#define sca2(a,b) scanf("%d%d",&a,&b );
#define sca3(a,b,c) scanf("%d%d%d",&a,&b,&c);
#define mem(a,b) memset(a,b,sizeof(a));
#define FOR0(i,j) for(int i=0;i<=j;++i)
#define FOR1(i,j) for(int i=1;i<=j;++i)
using namespace std;
struct node
{
    int x,y,z,step;
};
char plat[40][40][40];
int vis[40][40][40];
int dir[6][3]={0,0,1,0,1,0,1,0,0,-1,0,0,0,-1,0,0,0,-1};
int m,n,o;
int sx,sy,sz;
void ini()
{
    mem(plat,0);
    mem(vis,0);
}
int main()
{
    //cout<<sx<<sy<<sz;
    while(scanf("%d%d%d",&o,&m,&n)&&m+n+o)
    {
        ini();
        for(int i=1;i<=o;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%s", plat[i][j]+1);
                for(int k=1;k<=n;k++)
                {
                    if(plat[i][j][k]=='S')
                    {
                        sx=j;sy=k;sz=i;
                    }
                }
            }
        }
        //cout<<sx<<sy<<sz;
        queue<node>q;
        q.push({sx,sy,sz,0});
        vis[sz][sx][sy]=1;
        int flag=0;
        while(!q.empty())
        {
            node now=q.front();
            q.pop();
            if(plat[now.z][now.x][now.y]=='E')
            {
                flag=1;
                printf("Escaped in %d minute(s).
", now.step);
                break;
            }
            int nex,ney,nez;
            for(int i=0;i<6;i++)
            {
                nex=now.x+dir[i][0];
                ney=now.y+dir[i][1];
                nez=now.z+dir[i][2];
                if(nex<1||nex>m||ney<1||ney>n||nez<1||nez>o) continue;
                if(vis[nez][nex][ney]) continue;
                if(plat[nez][nex][ney]=='.'||plat[nez][nex][ney]=='E')
                {
                    //printf("%d %d %d
",nex,ney,nez );
                    vis[nez][nex][ney]=1;
                    q.push({nex,ney,nez,now.step+1});
                }
            }
        }
        if(!flag)
            printf("Trapped!
");
}
return 0;

}