POJ 2488A Knight's Journey(搜索入门)

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分类专栏：北大acm 搜索文章标签： POJ 2488A Knights J

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探讨了骑士周游问题的解决方法，即如何使国际象棋中的骑士遍历特定大小的棋盘上的每一个格子且不重复。通过深度优先搜索算法找到了字典序最小的解决方案。

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A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46861		Accepted: 15959

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

题意：给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

想法：深搜

代码：

#include <iostream>
#include<stdio.h>
#include <cstring>
using namespace std;
const int N = 35;
int vis[N][N];
int p, q, flag;
char b[500];
int dir[8][2] = {-1, -2, 1, -2, -2, -1, 2, -1, -2, 1, 2, 1, -1, 2, 1, 2};
void dfs(int x, int y, int cnt)
{
    if (cnt == p*q)
    {
        printf("A1");
        for (int i = 2; i < 2*cnt; i++)
           printf("%c",b[i]);
        printf("\n\n");
        flag = 1;
        return;
    }
    for (int i = 0; i < 8 && !flag; i++)
    {
       int tx = x + dir[i][0], ty = y + dir[i][1];
       if (tx < 1 || tx > p || ty < 1 || ty > q)
        continue;
       if (!vis[tx][ty])
       {
           vis[tx][ty] = 1;
           b[2*cnt] = 'A' + ty - 1;
           b[2*cnt+1] = '1' + tx - 1;
           dfs(tx, ty, cnt+1);
           vis[tx][ty] = 0;
        }
    }
}
int main()
{
    int t, ca = 0;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d %d",&p,&q);
        printf("Scenario #%d:\n",++ca);
        memset(vis, 0, sizeof(vis));
        flag = 0;
        vis[1][1] = 1;
        dfs(1, 1, 1);
        if (!flag) printf("impossible\n\n");
    }
    return 0;
}