poj2488题(骑士以马走日的方式走棋盘)

A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 40636 Accepted: 13807

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

3
1 1
2 3
4 3
Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题目意思是：从所给棋盘的第一个位置出发，深度优先搜索棋盘，但是每一次搜索都有八个方向，注意方向也得按照字典序，就是按照马走日的方法，不重复遍历，判断能不能把这个棋盘给遍历完成。博客里的代码大部分都是借鉴别人的，想把这些方法总结起来。

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<iostream>using namespace std;int visited[27][27];int p, q;//p指列指数字，q指行指字母int flag;int path[27][2];//记录路径int dir[8][2] = { -2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1 };//指八个方向void build(int a,int b){    int i, j;    for (i = 1;i <= b;i++)        for (j = 1;j <= a;j++)            visited[i][j] = 0;}void dfs(int a,int b,int step)//step记录步数{    if (flag)return;    path[step][0] = a;    path[step][1] = b;    if (step==p*q)    {        flag = 1;        return;    }    for (int i = 0;i < 8;i++)    {        int nx, ny;        nx = a + dir[i][0];        ny = b + dir[i][1];        if (nx >= 1 && nx <= q&&ny >= 1 && ny <= p&&!visited[nx][ny])        {            visited[nx][ny] = 1;            dfs(nx, ny, step + 1);            visited[nx][ny]=0;//如果上一次搜索未能成功，返回到上一步，把该处的状态返回原样        }    }    return;}int main(){    int n,i;    cin >> n;    for (i = 1;i <= n;i++)    {        cin >> p;        cin >> q;        build(p,q);        memset(path, 0, sizeof(path));        flag = 0;        visited[1][1] = 1;        dfs(1,1,1);        cout << "Scenario #" << i <<":"<< endl;        if (flag)        {            for (int j = 1;j <= p*q;j++)                printf("%c%d", path[j][0] - 1 + 'A', path[j][1]);            printf("\n");        }        else            printf("impossible\n");        if(i!=n)            printf("\n");    }}