POJ 2488 DFS

Posted 2020-10-17 liylho

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：

 1 //#include "bits/stdc++.h"
 2 #include "cstdio"
 3 #include "map"
 4 #include "set"
 5 #include "cmath"
 6 #include "queue"
 7 #include "vector"
 8 #include "string"
 9 #include "cstring"
10 #include "time.h"
11 #include "iostream"
12 #include "stdlib.h"
13 #include "algorithm"
14 #define db double
15 #define ll long long
16 #define vec vector<ll>
17 #define Mt  vector<vec>
18 #define ci(x) scanf("%d",&x)
19 #define cd(x) scanf("%lf",&x)
20 #define cl(x) scanf("%lld",&x)
21 #define pi(x) printf("%d\n",x)
22 #define pd(x) printf("%f\n",x)
23 #define pl(x) printf("%lld\n",x)
24 #define rep(i, x, y) for(int i=x;i<=y;i++)
25 const int N   = 1e6 + 5;
26 const int mod = 1e9 + 7;
27 const int MOD = mod - 1;
28 const db  eps = 1e-18;
29 const db  PI  = acos(-1.0);
30 using namespace std;
31 int n,m,ok=0;
32 int dx[8]={-2,-2,-1,-1,1,1,2,2};
33 int dy[8]={-1,1,-2,2,-2,2,-1,1};
34 bool v[30][30];
35 struct P
36 {
37     int x,y;
38 };
39 P a[N];
40 int R()
41 {
42     int x=0,f=1;char ch=getchar();
43     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
44     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
45     return x*f;
46 }
47 void  dfs(int x,int y,int cnt,int ii)
48 {
49     v[x][y]=1;//（1）
50     a[cnt].x=x,a[cnt].y=y;//（1）
51     if(cnt==n*m&&!ok)
52     {
53         ok=1;
54         printf("Scenario #%d:\n",ii);
55         for(int i=1;i<=cnt;i++){
56             char e=(a[i].x+‘A‘-1);
57             printf("%c%d",e,a[i].y);
58         }
59         puts("");
60         puts("");
61         return;
62     }
63     for(int i=0;i<8;i++)
64     {
65         int xx=x+dx[i],yy=y+dy[i];
66         if(xx<=0||yy<=0||xx>m||yy>n||v[xx][yy]!=0) continue;
67         /*------（1)不能放在这里?------*/
68 
69         dfs(xx,yy,cnt+1,ii);
70         v[xx][yy]=0;
71     }
72 }
73 int main()
74 {
75     int t;
76     t=R();
77     for(int ii=1;ii<=t;ii++)
78     {
79         memset(v,0, sizeof(v));
80         n=R();m=R();
81         ok=0;
82         dfs(1,1,1,ii);
83         if(!ok){
84             printf("Scenario #%d:\n",ii);
85             puts("impossible");
86             puts("");
87         }
88 
89     }
90     return 0;
91 }