POJ 2488 A Knight's Journey【DFS + 回溯应用】

原题链接：http://poj.org/problem?id=2488

我的链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=19651#problem/A

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23452 Accepted: 7944

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

算法：DFS + 回溯

题意：骑士周游列国问题，说白了就是任一给你一个棋盘（当然满足格子数小于26）

            让你从第一个格子按照图中所给的马走日的步法

            不重复遍历走完整个棋盘。然后按照字典序输出路径即可

思路：按照图中的八个方向dfs即可。

注意: 1.回溯的用法，千万别忘了还原状态。

         2.路径的输出，字典序的是列。。。 

 

/******************************************************//AAccepted148 KB0 msC++1259 B题意：骑士周游列国,马走日八个方向走,同时也要注意字典序      最后是按照字典序输出的      注意:先输入列再输入行*******************************************************/#include<stdio.h>#include<string.h>const int maxn = 30;int vis[maxn][maxn];int n,m;int dir[8][2] = {-2,-1, -2,1, -1,-2, -1,2, 1,-2, 1,2, 2,-1, 2,1};//注意方向也要按照字典序来 ,否则WA的好惨。。。struct Path{    int x,y;}p[maxn];int flag;void dfs(int x, int y, int step){    if(flag) return;    p[step].x = x;    p[step].y = y;    if(step == n*m)    {        flag = 1;        return;    }    Path next;    for(int i = 0; i < 8; i++)    {        next.x = x+dir[i][0];        next.y = y+dir[i][1];        if(next.x >= 1 && next.x <= n && next.y >= 1 && next.y <= m && !vis[next.x][next.y])        {            vis[next.x][next.y] = 1;            dfs(next.x, next.y, step+1);            vis[next.x][next.y] = 0;        }    }    return;}int main(){    int T;    scanf("%d", &T);    for(int t = 1; t <= T; t++)    {        scanf("%d%d", &m,&n); //先输入列再输入行        memset(vis,0,sizeof(vis));        memset(p,0,sizeof(p));        flag = 0;        vis[1][1] = 1;        dfs(1,1,1);        printf("Scenario #%d:\n", t);        if(flag)        {            for(int i = 1; i <= n*m; i++)                printf("%c%d", p[i].x-1+'A', p[i].y);            printf("\n");        }        else printf("impossible\n");        if(t != T) printf("\n");    }    return 0;}