POJ 2488-A Knight's Journey（DFS） - c++编程基础

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 31702		Accepted: 10813

Description

Background
The knight is getting bZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"pst">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

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

题意：国际象棋，然后给一个马（马走日） ，可以从任意点出发，找一条可以访问所有格子（p*q的棋盘）的路径，注意路径如果有多条要求输出字典序最小的那条。。然后这个可以搜索的时候按字典序搜。。就是搜索方向要固定。。不能随意写了

然后其他的没什么了 直接深搜，搜到答案之后直接return ；

#include 
   
    
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                 #define ll long long #define maxn 116 #define pp pair
                
                  #define INF 0x3f3f3f3f #define max(x,y) ( ((x) > (y)) ? (x) : (y) ) #define min(x,y) ( ((x) > (y)) ? (y) : (x) ) using namespace std; int n,m,k,ans,dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}; bool vis[27][27]; int sx[30],sy[30],top,ok; void dfs(int x,int y) { if(ok) return ; if(top==n*m) { ok=1; for(int i=0;i
                 
                  =1&&tx<=n&&ty>=1&&ty<=m&&!vis[tx][ty]) { vis[tx][ty]=1;sx[top]=tx;sy[top++]=ty; dfs(tx,ty); vis[tx][ty]=0;top--; } } } int main() { int T,cas=1; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m);ok=0; printf("Scenario #%d:\n",cas++); memset(vis,0,sizeof(vis));top=0; vis[1][1]=1;sx[top]=1;sy[top++]=1; dfs(1,1); if(!ok) printf("impossible"); puts("");if(T)puts(""); } return 0; }