A Knight's Journey（DFS）

A - A Knight's Journey

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice POJ 2488

Appoint description:System Crawler (2016-03-21)

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

题意：

    寻找一条能将所有的格子走完的路径，是按照中国象棋的马的走法。

思路：

    这题不难，直接dfs暴力一波就行了。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<vector>using namespace std;#define CRL(a) memset(a,0,sizeof(a))vector< pair<int,int> > k;pair<int,int> v;int n,m,flag,cnt;int fx[][2] = {{-1,-2},{1,-2},{-2,-1},{2,-1}             ,{-2,1},{2,1},{-1,2},{1,2}};int map[55][55];bool jugde(int x,int y){if(x>=1&&x<=n&&y>=1&&y<=m&&!map[x][y]){return true;}return false;}void dfs(int x,int y,int c){int tx,ty;if(c==n*m){flag=0;}if(!flag)return;for(int i=0;i<8;++i){v.first=x, tx = fx[i][0] + x;v.first=tx;v.second=y, ty = fx[i][1] + y;v.second=ty;if(jugde(tx,ty)){k.push_back(v);map[tx][ty]=1;dfs(tx,ty,c+1);if(!flag)return;  k.pop_back();  map[tx][ty]=0;}}}int main(){/*freopen("input.txt","r",stdin);*/int N,i,j,c=0;scanf("%d",&N);while(N--){flag = 1;cnt=0;scanf("%d%d",&n,&m);printf("Scenario #%d:\n",++c);for(i=1;i<=m&&flag;++i){for(j=1;j<=n&&flag;++j){map[j][i] = 1;v.first = j,v.second = i;k.push_back(v);dfs(j,i,1);CRL(map);if(flag)k.clear();   }}if(!flag){for(i=0;i<k.size();++i){printf("%c%d",(k[i].second+'A'-1),k[i].first);}printf("\n");k.clear();}else{printf("impossible\n");}printf("\n");}return 0;}