POJ 2488（DFS）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal 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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #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 30int t, n, m;int g[N][N];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[N];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 CDZSCfreopen("i.txt", "r", stdin);#endifint 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;}