poj 2488A Knight's Journey（DFS＋回溯）

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 46464 Accepted: 15812

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

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#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;using namespace std;const int maxn=30;int vis[maxn][maxn];int n,m;int dx[8]={-2,-2,-1,-1,1,1,2,2};int dy[8]={-1,1,-2,2,-2,2,-1,1};//方向要按照字典序struct node{    int x,y;}p[maxn];int flag;void dfs(int x,int y,int step){    if(flag) return ;    p[step].x=x;    p[step].y=y;    if(step==n*m)    {        flag=1;        return;    }    node a;    for(int i=0;i<8;i++)    {        a.x=x+dx[i];        a.y=y+dy[i];        if(a.x>0 && a.x<=n && a.y>0 && a.y<=m && !vis[a.x][a.y])        {            vis[a.x][a.y]=1;            dfs(a.x,a.y,step+1);            vis[a.x][a.y]=0;        }    }}int main(){    int T;    cin>>T;    int cas=1;    while(T--)    {        cin>>m>>n;        memset(vis,0,sizeof(vis));        memset(p,0,sizeof(p));        vis[1][1]=1;        flag=0;        dfs(1,1,1);        cout<<"Scenario #"<<cas++<<":"<<endl;        if(flag)        {            for(int i=1;i<=n*m;i++)                printf("%c%d",p[i].x-1+'A',p[i].y);                //cout<<(char)(p[i].x-1-+'A')<<p[i].y;            cout<<endl;        }        else        {            cout<<"impossible"<<endl;        }        if(T) cout<<endl;    }    return 0;}