poj 2488 A Knight's Journey 简单dfs回溯 字典序

http://poj.org/problem?id=2488题目链接

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29706 Accepted: 10173

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

题意是一个马在p*q的棋盘内，由一点开始跳，能否跳满整块棋盘 并输出字典序最小的路径

看到需要字典序最小 于是就从1,1开始试了一下结果直接ac了 据说能用反证法证明 如果1,1impossible 则其它一定impossible

需要注意的是dx和dy的顺序

代码如下：

#include<cstdio>#include<cstring>#include<stdlib.h>#include<queue>#include<vector>#include<iostream>#include<algorithm>using namespace std;bool map[105][105];char a[60];int dx[10]={-1,1,-2,2,-2,2,-1,1};int dy[10]={-2,-2,-1,-1,1,1,2,2};int p,q;int dfs(int x,int y,int step){    if(step==p*q)        return 1;    int nx,ny,i;    map[x][y]=1;    for(i=0;i<8;i++)    {        nx=x+dx[i];        ny=y+dy[i];        a[step*2]='A'-1+ny;        a[step*2+1]='0'+nx;        if(nx>p||ny>q||nx<1||ny<1||map[nx][ny])        continue;        if(dfs(nx,ny,step+1))            return 1;        map[nx][ny]=0;    }    return 0;}int main(){    int n,i;    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d%d",&p,&q);            memset(map,0,sizeof(map));            memset(a,0,sizeof(a));            a[0]='A';            a[1]='1';            printf("Scenario #%d:\n",i);            if(dfs(1,1,1))                printf("%s\n\n",a);            else printf("impossible\n\n");        }    }    return 0;}