A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 24840 Accepted: 8412
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 3Sample Output
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
Source
簡單的深搜,不多說,直接上代碼!
#include<iostream> #include<stdio.h> #include<cstring> using namespace std; int pathlow[30],pathdown[30],visit[30][30]; int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}},n,m;//這裡注意是字典序最小 bool dfs(int low,int down,int num) { int i,x,y; if(num==n*m) { for(i=0;i<n*m;i++) { printf("%c%d",'A'+pathdown[i],pathlow[i]+1); } return true; } for(i=0;i<8;i++) { x=low+dir[i][0]; y=down+dir[i][1]; pathlow[num]=x; pathdown[num]=y; if(x>=0&&x<n&&y>=0&&y<m&&(!visit[x][y])) { visit[x][y]=1; if(dfs(x,y,num+1)) { return true; } else { visit[x][y]=0;//這裡要注意,一定要重新標記為0 } } } return false; } int main () { int t,i; while(scanf("%d",&t)!=EOF) { for(i=1;i<=t;i++) { printf("Scenario #%d:\n",i); scanf("%d%d",&n,&m); memset(visit,0,sizeof(visit)); visit[0][0]=1; pathlow[0]=0; pathdown[0]=0; if(!dfs(0,0,1)) { printf("impossible"); } printf("\n\n"); } } return 0; } #include<iostream> #include<stdio.h> #include<cstring> using namespace std; int pathlow[30],pathdown[30],visit[30][30]; int dir[8][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}},n,m;//這裡注意是字典序最小 bool dfs(int low,int down,int num) { int i,x,y; if(num==n*m) { for(i=0;i<n*m;i++) { printf("%c%d",'A'+pathdown[i],pathlow[i]+1); } return true; } for(i=0;i<8;i++) { x=low+dir[i][0]; y=down+dir[i][1]; pathlow[num]=x; pathdown[num]=y; if(x>=0&&x<n&&y>=0&&y<m&&(!visit[x][y])) { visit[x][y]=1; if(dfs(x,y,num+1)) { return true; } else { visit[x][y]=0;//這裡要注意,一定要重新標記為0 } } } return false; } int main () { int t,i; while(scanf("%d",&t)!=EOF) { for(i=1;i<=t;i++) { printf("Scenario #%d:\n",i); scanf("%d%d",&n,&m); memset(visit,0,sizeof(visit)); visit[0][0]=1; pathlow[0]=0; pathdown[0]=0; if(!dfs(0,0,1)) { printf("impossible"); } printf("\n\n"); } } return 0; }