POJ 2488 DFS

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47611		Accepted: 16183

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

代码：

//#include "bits/stdc++.h"
#include "cstdio"
#include "map"
#include "set"
#include "cmath"
#include "queue"
#include "vector"
#include "string"
#include "cstring"
#include "time.h"
#include "iostream"
#include "stdlib.h"
#include "algorithm"
#define db double
#define ll long long
#define vec vector<ll>
#define Mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i, x, y) for(int i=x;i<=y;i++)
const int N   = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const db  eps = 1e-18;
const db  PI  = acos(-1.0);
using namespace std;
int n,m,ok=0;
int dx[8]={-2,-2,-1,-1,1,1,2,2};
int dy[8]={-1,1,-2,2,-2,2,-1,1};
bool v[30][30];
struct P
{
    int x,y;
};
P a[N];
int R()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void  dfs(int x,int y,int cnt,int ii)
{
    v[x][y]=1;//（1）
    a[cnt].x=x,a[cnt].y=y;//（1）
    if(cnt==n*m&&!ok)
    {
        ok=1;
        printf("Scenario #%d:
",ii);
        for(int i=1;i<=cnt;i++){
            char e=(a[i].x+'A'-1);
            printf("%c%d",e,a[i].y);
        }
        puts("");
        puts("");
        return;
    }
    for(int i=0;i<8;i++)
    {
        int xx=x+dx[i],yy=y+dy[i];
        if(xx<=0||yy<=0||xx>m||yy>n||v[xx][yy]!=0) continue;
        /*------（1)不能放在这里?------*/

        dfs(xx,yy,cnt+1,ii);
        v[xx][yy]=0;
    }
}
int main()
{
    int t;
    t=R();
    for(int ii=1;ii<=t;ii++)
    {
        memset(v,0, sizeof(v));
        n=R();m=R();
        ok=0;
        dfs(1,1,1,ii);
        if(!ok){
            printf("Scenario #%d:
",ii);
            puts("impossible");
            puts("");
        }

    }
    return 0;
}