POJ2488-A Knight's Journey（DFS）_poj 2488 a knight's journey-优快云博客

POJ2488 原题链接:http://poj.org/problem?id=2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46885		Accepted: 15967

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

题目大意:给一个p*q的国际象棋棋盘,棋盘上有马从任意起点开始,按照马的跳法跳完整个棋盘,如果能则输出最小字典序的路径,如果不能则输出impossible。

思路:

1.看到这道题很容易想到dfs,这道题是判断是否能够一遍走完,如何判断是否走完呢？可以添加一个步数数据与一个标记,若无路可走时步数恰好等于格子数,则说明可以一次走完,标记为true,若走到最后无路可走时步数不等于格子数,那么说明这种走法不能一次走完,标记为false。

2.如果不重复走一遍就走完了，标记为true，算法停止,假若在某种dfs下走到某一步时无路可走而棋盘还有未走到的点(步数小于格子数），那么退回这一步，尝试其他的路线，退回这一步就是在递归深搜返回时重置该点，以便在当前路线走一遍行不通换另一种路线时，该点的状态是未访问过的，而不是像普通的dfs当作已经访问了。

3.题目要求输出的路径为字典序最小的路径,所以我们在跳马的时候的优先顺序应以字典序最小的顺序,且从A1点开始走。

代码如下：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
using namespace std;
const int MAX_N = 27;
int p,q;
struct Step{
    char x,y;
}path[MAX_N];
bool zouwan;//标记是否已经跳完
bool vis[MAX_N][MAX_N];
const int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
const int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
void DFS(int x,int y,int step)
{
    int vx,vy;
    path[step].x=x+'0';//int 转为 char
    path[step].y=y+'A'-1;
    if(step==p*q)//步数等于格子数说明可以一次跳完
    {
        zouwan=true;//标记为可以跳完
        return;
    }
    for(int i=0;i<8;i++)
    {
        vx=x+dx[i];
        vy=y+dy[i];
        if (0<vx&&vx<=p&&0<vy&&vy<=q&&!vis[vx][vy]&&!zouwan)
        {
            vis[vx][vy] = true;
            DFS(vx,vy,step+1);
            vis[vx][vy] = false;//退回该步
        }
    }

}
int main()
{
    int T;
    scanf("%d",&T);
    for(int c=1;c<=T;c++)
    {
        zouwan=false;
        scanf("%d %d",&p,&q);
        int num=0;//步数标记
        memset(vis,false,sizeof(vis));
        vis[1][1]=true;
        DFS(1,1,1);
        printf("Scenario #%d:\n", c);
        if(zouwan)//可以跳完输出路径
        {
            for(int i=1;i<=p*q;i++)
            {
                printf("%c%c",path[i].y,path[i].x);
            }
            printf("\n");
        }
        else printf("impossible\n");//不能跳完输出impossible
        if (c != T)
            printf("\n");
    }
}