快速切题 poj2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31195		Accepted: 10668

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

实际用时 20min
情况:CCCCA 注意java和胡乱改动
注意点:1 组间空行但最后一组没有 2 字典序

#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
typedef unsigned long long ull;
bool used[8][8];
char heap[64][3];
const int dx[8]={-2,-2,-1,-1,1,1,2,2},dy[8]={-1,1,-2,2,-2,2,-1,1};
bool judge(int x,int y){
    if(x>=0&&x<n&&y>=0&&y<m)return true;
    return false;
}
bool dfs(int x,int y,int cnt){
    used[x][y]=true;
    heap[cnt][0]=x+'A';
    heap[cnt++][1]=y+'1';
    if(cnt==n*m)return true;
    for(int i=0;i<8;i++){
        int tx=x+dx[i],ty=y+dy[i];
        if(judge(tx,ty)&&!used[tx][ty]){
            if(dfs(tx,ty,cnt))return true;
        }
    }
    used[x][y]=false;
    return false;
}
int main(){
    int T;scanf("%d",&T);
    for(int ti=1;ti<=T;ti++){
        scanf("%d%d",&m,&n);
        memset(used,0,sizeof(used));
        bool fl=dfs(0,0,0);
        printf("Scenario #%d:\n",ti);
        if(fl){
            for(int i=0;i<n*m;i++){
                printf("%s",heap[i]);
            }
            puts("");
        }
        else {
            puts("impossible");
        }
        if(ti<T)puts("");
    }
    return 0;
}