[poj]2488 A Knight's Journey dfs+路径打印
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 45941 | Accepted: 15637 |
Description

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
Output
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
按照字典序输出路径,方向要按照字典序来搜索。
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
bool v[][];
int p, q;
int dir[][] = {{-,-},{-,},{-,-},{-,},
{,-},{,},{,-},{,}};
int px[], py[];
int step, flag;
char R[] = {'A','B','C','D','E','F','G','H'}; int dfs(int x, int y, int step)
{
if (step == p*q) {
flag = ;
for (int i = ; i < p*q; i++) {
printf("%c%d", R[px[i]],py[i]+);
}
printf("\n\n");
return ;
} int nx, ny;
for (int i = ; i < ; i++) {
nx = x + dir[i][];
ny = y + dir[i][];
if (!v[nx][ny] && nx>= && nx<q && ny>= && ny<p) {
v[nx][ny] = ;
px[step] = nx; py[step] = ny;
dfs(nx, ny, step+);
if (flag) return ; //只搜索一次
v[nx][ny] = ;
}
}
return ;
} int main()
{
//freopen("1.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int T;
int t = ;
cin >> T;
while (T--) {
cin >> p >> q;
printf("Scenario #%d:\n", ++t);
memset(v, , sizeof(v));
px[] = ; py[] = ;
v[][] = ;
flag = ;
step = ;
if(!dfs(, , ))
printf("impossible\n\n");
} return ;
}
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