POJ 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2488
64-bit integer IO format: %lld      Java class name: Main

 

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

 

 

解题：搜索。图中显示了移动方向，注意字典序小到大。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
int n,m,path[100][2];
bool vis[100][100];
bool dfs(int x,int y,int cur){
    if(cur >= n*m) return true;
    for(int i = 0; i < 8; i++){
        int tx = dir[i][0]+x;
        int ty = dir[i][1]+y;
        if(tx < 0 || tx >= n || ty < 0 || ty >= m || vis[tx][ty]) continue;
        vis[tx][ty] = true;
        path[cur][0] = tx+'A';
        path[cur][1] = ty+1;
        if(dfs(tx,ty,cur+1)) return true;
        vis[tx][ty] = false;
    }
    return false;
}
int main() {
    int t,k = 1;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&m,&n);
        bool flag = false;
        memset(vis,false,sizeof(vis));
        path[0][0] = 'A';
        path[0][1] = 1;
        vis[0][0] = true;
        printf("Scenario #%d:
",k++);
        if(dfs(0,0,1)){
            for(int i = 0; i < n*m; i++) printf("%c%d",path[i][0],path[i][1]);
            puts("");
        }else puts("impossible");
        puts("");
    }
    return 0;
}