POJ 2488 A Knight's Journey(DFS+回溯)

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46639		Accepted: 15875

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

题目大意：给你N组数据，输入N组p,q，p代表棋盘的行数，q代表棋盘的列数。任选一个起点，按照国际象棋马的跳法，不重复的跳完整个棋盘，如果有多种路线则选择字典序最小的路线（路线是点的横纵坐标的集合，注意棋盘的横坐标的用大写字母，纵坐标是数字)

题目分析：

1. 应该看到这个题就可以想到用DFS，当首先要明白这个题的意思是能否只走一遍（不回头不重复）将整个地图走完，而普通的深度优先搜索是一直走，走不通之后沿路返回到某处继续深搜。所以这个题要用到的回溯思想，如果不重复走一遍就走完了，做一个标记，算法停止；否则在某种DFS下走到某一步时按马跳的规则无路可走而棋盘还有为走到的点，这样我们就需要撤消这一步，进而尝试其他的路线（当然其他的路线也可能导致撤销），而所谓撤销这一步就是在递归深搜返回时重置该点，以便在当前路线走一遍行不通换另一种路线时，该点的状态是未访问过的，而不是像普通的DFS当作已经访问了。

2. 如果有多种方式可以不重复走一遍的走完，需要输出按字典序最小的路径，而注意到国际象棋的棋盘是列为字母，行为数字，如果能够不回头走一遍的走完，一定会经过A1点，所以我们应该从A1开始搜索，以确保之后得到的路径字典序是最小的（也就是说如果路径不以A1开始，该路径一定不是字典序最小路径），而且我们应该确保优先选择的方向是字典序最小的方向，这样我们最先得到的路径就是字典序最小的。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int vis[27][27];
int dir[9][2]={0,0,-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2};
int x,y;
bool flag=false;
struct node
{
    int x,y;
}path[27];
bool check(int x1,int y1)
{
    if(x1<1||x1>x||y1<1||y1>y||flag==true||vis[x1][y1]==1) return false;
    else return true;
}
void DFS(int x0,int y0,int num)
{
    int i;
    path[num].x=x0,path[num].y=y0;
    //printf("x0y0:%d %d %d\n",x0,y0,vis[x0][y0]);
    if(num==x*y)  {flag=true;return ;}
    for(i=1;i<=8;i++)
    {
        int x1=x0+dir[i][0],y1=y0+dir[i][1];
        if(check(x1,y1))
        {
            vis[x1][y1]=1;
            DFS(x1,y1,num+1);
            vis[x1][y1]=0;//回溯
        }
    }
}
int main()
{
    int n,i;
    char c;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        flag=false;
        memset(vis,0,sizeof(vis));
        scanf("%d %d",&x,&y);
        vis[1][1]=1;
        DFS(1,1,1);

        if(flag==false) printf("Scenario #%d:\nimpossible\n",i);
        else
        {
            printf("Scenario #%d:\n",i);
            for(int j=1;j<=x*y;j++)
            {
                printf("%c%d",path[j].y+64,path[j].x);
            }
            printf("\n");
        }
        if(i!=n) printf("\n");
    }
    return 0;
}

原文链接：https://blog.csdn.net/anan15151529/article/details/77119543