POJ 2488（DFS）

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 38555		Accepted: 13076

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

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意：有个骑士在点（0,0），问它能否不重复的走完所有的点，如果可以就输出字典序最小的路径，否则就输出impossible

题解：直接暴力DFS就好啦，暴力出所有的路径，按照字典序排序

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
using namespace std;
#define N 30
int t, n, m;
int g

;
bool flag;
int vx[] = { -1,1,-2,2,-1,1,-2,2 };
int vy[] = { 2,2,1,1,-2,-2,-1,-1 };
bool judge(int x, int y)
{
if (x < 0 || x >= n || y<0 || y >= m)return 0;
if (g[x][y])return 0;
return 1;
}
pair<int, int>path
;
set<string>ans;
void dfs(int x, int y, int stp)
{
if (stp == n*m)
{
string tmp;
flag = true;
for (int i = 0; i < stp; i++)
{

tmp += char(path[i].second + 'A');
tmp += char(path[i].first + '0'+1);
}
ans.insert(tmp);
return;
}
for (int i = 0; i < 8; i++)
{
int xx = x + vx[i];
int yy = y + vy[i];
if (judge(xx, yy))
{
g[xx][yy] = 1;
path[stp].first = xx;
path[stp].second = yy;
dfs(xx, yy, stp + 1);
g[xx][yy] = 0;
}
}
}
int main()
{
#ifdef CDZSC
freopen("i.txt", "r", stdin);
#endif
int cas = 0;
scanf("%d", &t);
while (t--)
{
ans.clear();
if (cas)puts("");
printf("Scenario #%d:\n", ++cas);
flag = false;
memset(g, 0, sizeof(g));
scanf("%d%d", &n, &m);
path[0].first = 0;
path[0].second = 0;
g[0][0] = 1;
dfs(0, 0, 1);
if (!flag)puts("impossible");
else
{
printf("%s\n", ans.begin()->c_str());
}
}
return 0;
}