POJ2488 A Knight's Journey[dfs]_d - a knight's journey-优快云博客

本文链接：https://blog.youkuaiyun.com/ControlBear/article/details/51838001

本文探讨了一个经典的计算机科学问题——骑士周游问题。该问题要求在一个缩小版的国际象棋棋盘上找出一条路径，使得骑士能够按照国际象棋中骑士的移动规则访问每个格子恰好一次。文章提供了一种通过深度优先搜索来寻找解决方案的方法，并附带了完整的C++实现代码。

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D - A Knight's Journey

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

题意:

给定 n,m 看马能不能走遍整个棋盘按字典序输出

简单的dfs 注意了next的写法因为要按字典序输出

其实找到应该就可以直接返回了,没直接返回感觉更耗时

之后写的时候注意一下就好.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
const int N=30;
bool mark[N][N],flag;
int next[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int road[N*N][2];
int p,q;
void dfs(int x,int y,int step)
{
    road[step][0]=x;
    road[step][1]=y;
    if (step==p*q)
    {
        flag=true;
        return ;
    }
    for (int i=0;i<8;i++)
    {
        int nx=x+next[i][0];
        int ny=y+next[i][1];
        if (nx>0 && nx<=p && ny>0 && ny<=q && !mark[nx][ny] && !flag)//记得回溯
        {
            mark[nx][ny]=1;
            dfs(nx,ny,step+1);
            mark[nx][ny]=0;
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    for (int test=1;test<=T;test++)
    {
        memset(mark,0,sizeof(mark));
        scanf("%d%d",&p,&q);
        flag=0;
        mark[1][1]=1;
        dfs(1,1,1);
        printf("Scenario #%d:\n",test);
        if (flag)
            for (int i=1;i<=p*q;i++)
                printf("%c%d",road[i][1]+'A'-1,road[i][0]);//貌似题目那里是x y反过来? 之前一直以为自己错了, 但是反过来跟样例一样
        else
            printf("impossible");
        printf("\n\n");
    }
    return 0;
}