poj 2488(简单dfs,注意方向)
A Knight's Journey
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 22306 | Accepted: 7546 |
Description
![](http://poj.org/images/2488_1.jpg)
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.
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
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define maxn 26 int p,q,ans,fg; int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//开始看的是网上的字典序方向,感觉和我的方向不一样,发现是人家思路和我不一样,我是按照dir数组里面来表示的搜索方向,是按照我的字典序 struct node { char a; int num; }path[maxn*maxn]; int vis[maxn][maxn]; void dfs(int x,int y,int ret) { int i,curX,curY; if(fg) return; if(vis[x][y]) return; vis[x][y]=1; for(i=0;i<8;i++) { if(ret==ans) { fg=1; for(i=1;i<=ans;i++) printf("%c%d",path[i].a,path[i].num); printf("\n"); return; } curX=x+dir[i][0]; curY=y+dir[i][1]; if(curX<1||curY<1||curX>p||curY>q||vis[curX][curY]) continue; path[ret+1].a='A'+curY-1; path[ret+1].num=curX; dfs(curX,curY,ret+1); vis[curX][curY]=0; } } int main() { int n; int t=1; scanf("%d\n",&n); while(n--) { scanf("%d%d",&p,&q); memset(vis,0,sizeof(vis)); ans=q*p; fg=0; printf("Scenario #%d:\n",t++); path[1].a='A'; path[1].num=1; dfs(1,1,1); if(!fg) printf("impossible\n"); if(n!=0) printf("\n"); } return 0; }