poj A Knight's Journey(DFS)(字典序)
2017-01-06 20:43
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A Knight's Journey
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
Sample Output
ps:
弄了一下午,坑死在了字典序上0.0
这一题的字典序:就是先按列排序,较小的在前。然后按行排序,也是较小的在前。
int diri[8]={-1,1,-2,2,-2,2,-1,1};
int dirj[8]={-2,-2,-1,-1,1,1,2,2};
也就是(借鉴大神的图)
另外,国际象棋的棋盘:行为数字p,列为字母q
代码:
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
ps:
弄了一下午,坑死在了字典序上0.0
这一题的字典序:就是先按列排序,较小的在前。然后按行排序,也是较小的在前。
int diri[8]={-1,1,-2,2,-2,2,-1,1};
int dirj[8]={-2,-2,-1,-1,1,1,2,2};
也就是(借鉴大神的图)
另外,国际象棋的棋盘:行为数字p,列为字母q
代码:
#include<stdio.h> #include<string.h> int n,m,flag; int vis[30][30]; int to[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};//方向 struct node { int x,y; } w[30]; int judge(int x,int y) { if(x>0&&x<=n&&y>0&&y<=m&&!vis[x][y]) return 1; return 0; } void dfs(int x,int y,int step) { w[step].x=x,w[step].y=y; if(step==n*m)//如果已经遍历了所有的点 { for(int i=1; i<=step; i++) printf("%c%d",w[i].y-1+'A',w[i].x); printf("\n"); flag=1;//标记已经找到答案 } if(flag) return ; for(int i=0; i<8; i++) { int xx=x+to[i][0]; int yy=y+to[i][1]; if(judge(xx,yy)) { vis[xx][yy]=1; dfs(xx,yy,step+1); vis[xx][yy]=0; } } if(flag) return ; } int main() { int t,cont=0; scanf("%d",&t); while(++cont<=t) { memset(vis,0,sizeof(vis)); flag=0; cf36 scanf("%d%d",&n,&m); printf("Scenario #%d:\n",cont); if(n>8||m>8) printf("impossible\n"); else { vis[1][1]=1; dfs(1,1,1); if(!flag) printf("impossible\n"); } printf("\n"); } return 0; }
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