【DFS+回溯】A Knight's Journey

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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

来源

TUD Programming Contest 2005, Darmstadt, Germany

OJ:OpenJudge

题目大意

    给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。（摘自http://blog.csdn.net/lyhvoyage/article/details/18355471）国际象棋的马的走法如图所示，马走的是日字，和中国象棋是一样，是跳着走，但是国际象的马没有蹩脚，图片上标注的圆是马可以走的地方。

分析

1、要你找一个不重复走完所有棋盘的路径，所以用DFS+回溯，尝试多条路径，如果该路径不能走完就回溯。

2、能找到走完的所有格子的路径，就输出字典序最小的，又因为可以在任意一个各子开始，所以肯定直接从A1开始走，然后把遍历的方向按字典序从小到大排，这样就能保证按此方案能走完所有棋盘的路径就是题目要求的路径。

参考代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct path{//记录路径
int r;
char c;
}p[900];
int r,c;//row,column
int flag=0;//表示是否找到要求的路径

///p*q大于等于1小于等于26,所以p,q也是大于等于小于等于26
int visited[30][30];//标记数组，避免重复

/*
按字典序从小到大的顺序排列方向，
就能保证遍历时是字典序最小的路径
*/
const int di[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
const int dj[8] = {-2, -2, -1, -1, 1, 1, 2, 2};

void dfs(int si,int sj,int step);
int main()
{
    int n,num=1;
    scanf("%d",&n);
    while(n){
        scanf("%d%d",&r,&c);
        memset(visited,0,sizeof(visited));
        visited[1][1]=1;//A1点直接初始化
        dfs(1,1,1);
        printf("Scenario #%d:\n",num);
        if(flag==1){
            for(int i=1;i<=r*c;i++){
            printf("%c%d",p[i].c,p[i].r);
        }
        printf("\n\n");
        }
        else{
            printf("impossible\n\n");
        }
        flag=0;
        n--;
        num++;
    }
    return 0;
}
void dfs(int si,int sj,int step){
    //记录路径
    p[step].r=si;
    p[step].c=‘A‘+sj-1;
    //一共r*c个点所以能走r*c步就是能走完所有格子
    if(step==r*c){
        flag=1;
        return;
    }
    //遍历该点的八个方向
    for(int i=0;i<8;i++){
            int sii=si+di[i];
            int sjj=sj+dj[i];
            //必须满足这些条件才能进入函数
            if(sii>0&&sii<=r&&sjj>0&&sjj<=c&&visited[sii][sjj]==0&&flag==0){
            //进入函数的就标记下来
            visited[sii][sjj]=1;
            dfs(sii,sjj,step+1);
            //不能走完的路径会回溯回来，把这些走过但是不符合要求的点重新化0，让其他方向的路径试
            visited[sii][sjj]=0;
    }

    }

}

【DFS+回溯】A Knight's Journey

原文：http://www.cnblogs.com/LuRenJiang/p/7409836.html