POJ 2488:A Knight's Journey 深搜入门之走马观花
2015-07-31 14:57
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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
给一个棋盘,给你一匹马,马只能走“日”字格,问马能否不重复地走完所有格并且输出其走的位置的字典顺序。
头一次做深搜,中间出了不少问题,包括color在深搜之后居然没给它还原回去,这样的错误也行。。。
然后字典顺序是一个,如何记录结果path路径。
深搜给我的第一个感受就是能不用全局变量就不用全局变量,总在这里容易出错,另外,记录走的步数要设立一个参数step,之后再每层加1,这样的做法才是正途。。。
稀里糊涂地说了这么多,都是自己当时犯下的错,各种辛酸只有自己明白了。。。
代码:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 35342 | Accepted: 12051 |
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
给一个棋盘,给你一匹马,马只能走“日”字格,问马能否不重复地走完所有格并且输出其走的位置的字典顺序。
头一次做深搜,中间出了不少问题,包括color在深搜之后居然没给它还原回去,这样的错误也行。。。
然后字典顺序是一个,如何记录结果path路径。
深搜给我的第一个感受就是能不用全局变量就不用全局变量,总在这里容易出错,另外,记录走的步数要设立一个参数step,之后再每层加1,这样的做法才是正途。。。
稀里糊涂地说了这么多,都是自己当时犯下的错,各种辛酸只有自己明白了。。。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; struct struct_a{ int x; int y; }path[900]; int Test,i,p,q,m,n,flag,num; int move_x[10]={-1,1,-2,2,-2,2,-1,1};//改成字典顺序 int move_y[10]={-2,-2,-1,-1,1,1,2,2}; int color[30][30]; int dfs(int a,int b,int step) { if(flag==1) return 1; int k,temp_x,temp_y; if(step==q*p) { flag=1; return 1; } for(k=0;k<8;k++) { temp_x = a + move_x[k]; temp_y = b + move_y[k]; if(temp_x>=1 && temp_x<=p && temp_y>=1 && temp_y<=q && color[temp_x][temp_y]==0) { color[temp_x][temp_y] = color[a][b]+1; path[step].x=temp_x;//别用其他变量标记,递归这样不容易出现问题 path[step].y=temp_y; dfs(temp_x,temp_y,step+1); if(flag) return 1; } } color[a][b]=0;//如果最终选择不是这里,要记得清空 return 0; } int main() { cin>>Test; for(i=1;i<=Test;i++) { cin>>p>>q; cout<<"Scenario #"<<i<<":"<<endl; flag=0; memset(color,0,sizeof(color)); color[1][1]=1; path[0].x=1; path[0].y=1; dfs(1,1,1); if(flag==1) { int v; for(v=0;v<p*q;v++) { char temp_c=path[v].y-1+'A'; cout<<temp_c<<path[v].x; } cout<<endl; } else cout<<"impossible"<<endl; cout<<endl; } return 0; }
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