POJ 2488-A Knight's Journey(dfs)

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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

Output

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题意：给出一个规格小于8*8的棋盘，判断一个只能走“日”的骑士能否不重复的走遍整个棋盘，如果能，按字典序输出走的路径，否则输出“impossible”

思路：其实这道题唯一的难点就是按照字典序输出，只要解决了这个就是一个简单的dfs。其实在做很多题的时候我们在设置走的路径的时候都是随意设置的，这道题则不行，既然是按照字典序输出，则你设置路径的时候先按照列排序，让小的在前面，然后列的字母相同时，你按照行排序，同样让小的在前面。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>

using namespace std;
char map[30][2];
int vis[30][30];
int p,q;
int jx[]={-2,-2,-1,-1,1,1,2,2};
int jy[]={-1,1,-2,2,-2,2,-1,1};
int dfs(int x,int y,int cnt)
{
    int i;
    if(cnt==p*q){
        return 1;
    }
    for(i=0;i<8;i++){
        int dx=x+jx[i];
        int dy=y+jy[i];
        if(dx>=1&&dx<=q&&dy>=1&&dy<=p&&!vis[dx][dy]){
            map[cnt][0]=dx+'A'-1;
            map[cnt][1]=dy+'0';
            vis[dx][dy]=1;
            if(dfs(dx,dy,cnt+1))
                return 1;
            vis[dx][dy]=0;
        }
    }
    return 0;
}

int main()
{
    int n,i;
    int icase;
    scanf("%d",&n);
    for(icase=1;icase<=n;icase++){
        memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        scanf("%d %d",&p,&q);
        vis[1][1]=1;
        map[0][0]='A';
        map[0][1]='1';
        printf("Scenario #%d:\n",icase);
        if(dfs(1,1,1)){
            for(i=0;i<p*q;i++)
                printf("%c%c",map[i][0],map[i][1]);
        }
        else
            printf("impossible");
        printf("\n\n");
    }
    return 0;
}

POJ 2488-A Knight's Journey(dfs)

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原文地址：http://blog.csdn.net/u013486414/article/details/43876255