POJ-2488-A Knight's Journey

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#POJ-深度优先搜索

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本文探讨了一个经典的计算机科学问题——骑士周游问题。该问题要求在一个指定大小的国际象棋棋盘上找到一条路径，使得骑士能够访问每一个方格恰好一次，并提供了一种使用深度优先搜索算法来解决这一问题的方法。

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42499		Accepted: 14446

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

题目大意：给你一个n*m的棋盘问你骑士能否遍历棋盘上的所有点，如果可以按字典序遍历；

所以在DFS时搜索的次序要有先后

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int dx[8]={-1,1,-2,2,-2,2,-1,1};
int dy[8]={-2,-2,-1,-1,1,1,2,2};
int vist[27][27];

struct node
{
    int x,y;
}s[26*26];

int n,m,flag,cnt;

void DFS(int X,int Y)
{
    if(n*m == cnt)
    {
        flag = 1;
        return;
    }
    for(int l = 0; l < 8; l++)
    {
        int sx = X+dx[l];
        int sy = Y+dy[l];
        if(sx>=0 && sx<n && sy>=0 && sy<m && !vist[sx][sy])
        {
            s[cnt].x = sx;
            s[cnt++].y = sy;
            vist[sx][sy] = 1;
            DFS(sx,sy);
            if(!flag)
            {
                vist[sx][sy] = 0;
                cnt--;
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int i = 1; i <= t; i++)
    {
        scanf("%d%d",&n,&m);
        cnt = 1;
        flag = 0;
        memset(vist,0,sizeof(vist));
        vist[0][0] = 1;
        DFS(0,0);
        printf("Scenario #%d:\n",i);
        if(flag)
        {
            for(int j = 0; j < cnt; j++)
            {
                printf("%c%d",s[j].y+'A',s[j].x+1);
            }
            cout<<endl<<endl;
        }
        else
            cout<<"impossible"<<endl<<endl;
    }
}