Time limit1000 ms
Memory limit65536 kB
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
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4 题意：骑士走棋盘，要求把所有的各自都要走一遍，并且要输出走棋盘的格子题解：dfs搜索吧，注意每次可以搜索的时候都要把步数加一，当步数等于格子数时就可以了
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<stack>
using namespace std;
#define PI 3.14159265358979323846264338327950
int path[100][2],vis[100][100],p,q,cnt;
bool flag;
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
bool judge(int x,int y)
{
if(x<=p && x>=1 && y<=q && y>=1 && !vis[x][y] )
return true;
return false;
}
void dfs(int r,int c,int step)
{
if (flag == false)
{
path[step][0]=r;
path[step][1]=c;
}
if(step==p*q)
{
flag=true;
return ;
}
for(int i=0;i<8;i++)
{
int nx=r+dx[i];
int ny=c+dy[i];
if(judge(nx,ny))
{
vis[nx][ny]=1;
dfs(nx,ny,step+1);
vis[nx][ny]=0;
}
}
}
int main()
{
int i,t,cas=0;
cin>>t;
while(t--)
{
flag=0;
cin>>p>>q;
memset(vis,0,sizeof(vis));
vis[1][1]=1;
dfs(1,1,1);
printf("Scenario #%d:\n",++cas);
if(flag)
{
for(i=1;i<=p*q;i++)
{
printf("%c%d",path[i][1]-1+‘A‘,path[i][0]);
}
}
else
printf("impossible");
printf("\n");
if(t!=0)
printf("\n");
}
}
poj-2488 a knight's journey（搜索题）
原文地址：https://www.cnblogs.com/smallhester/p/9499143.html