HDU 2488 A Knight's Journey(dfs)-蒲公英云

HDU 2488 A Knight's Journey(dfs)

我就是我 2022-06-10 11:26 215阅读 0赞

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

题解：

题意：

给一个n*m的棋盘，和马走的八个方向，问你是否能走满整个棋盘，如果可以输出字典序最小的

思路：

这题以前在搜索专题做过，就是如果可以走满就直接从(1,1)这个点开始搜索，要保证字典序最小那么搜索的方向就要注意了，越右边的方向越优先考虑，找到一种就可以了，然后输出的时候先输出y再输出x这个有点坑

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
using namespace std;
#define lson k*2
#define rson k*2+1
#define M (t[k].l+t[k].r)/2
#define INF 1008611111
#define ll long long
#define eps 1e-15
int p[30][30];//存是否走过
int xx[30];//存路径的x坐标
int yy[30];
int dirx[8]={-1,1,-2,2,-2,2,-1,1};//方向比较重要
int diry[8]={-2,-2,-1,-1,1,1,2,2};
int n,m,tag;
void dfs(int x,int y,int step)
{
    if(step>=m*n)//如果走满了
    {
        tag=1;
    }
    if(tag)
    {
        return;
    }
    int i,j,sx,sy;
    for(i=0;i<8;i++)//八个方向dfs搜索回溯
    {
        sx=x+dirx[i];
        sy=y+diry[i];
        if(sx>=1&&sy>=1&&sx<=m&&sy<=n&&!p[sx][sy])
        {
            p[sx][sy]=1;
            xx[step]=sx;
            yy[step]=sy;//记录路径
            dfs(sx,sy,step+1);
            if(tag)
                return;
            p[sx][sy]=0;
        }
    }
}
int main()
{
    int i,j,test,q;
    scanf("%d",&test);
    for(q=1;q<=test;q++)
    {
        scanf("%d%d",&m,&n);
        memset(p,0,sizeof(p));
        tag=0;
        xx[0]=1;
        yy[0]=1;
        p[1][1]=1;
        dfs(1,1,1);//起点为1,1
        printf("Scenario #%d:\n",q);
        if(tag)
        for(i=0;i<n*m;i++)
        {
            printf("%c%d",'A'+yy[i]-1,xx[i]);//输出是先y再x
        }
        else
            printf("impossible");
        printf("\n");
        if(q!=test)//注意格式
            printf("\n");
    }
    return 0;
}