|
Problem Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part. Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b. |
|
Input
The input file will contain one or more test cases. Each test
case consists of one line containing two squares separated by one space.
A square is a string consisting of a letter (a-h) representing the
column and a digit (1-8) representing the row on the chessboard.
|
|
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.". |
|
Sample Input
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6 |
|
Sample Output
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves. |
|
Idea Well, I really have no word for this question. I use the common BFS, then it accepted. I think I can improve the algorithm by trimming, at least it have no need to search back. |
|
Code  View Code
1 #include <stdio.h> 2 #include <math.h> 3 #include <queue> 4 using namespace std; 5 struct Node 6 { 7 int x; 8 int y; 9 int step; 10 }; 11 int main() 12 { 13 int dx[10] = {2, 2, -2, -2, 1, 1, -1, -1}; 14 int dy[10] = {1, -1, 1, -1, 2, -2, 2, -2}; 15 char begin[5], end[5]; 16 int i, result; 17 Node beginN, endN, pre, cur; 18 while (scanf("%s %s", begin, end) != EOF) 19 { 20 queue<Node> q; 21 //translate the input 22 beginN.x = begin[0] - 'a'; 23 beginN.y = begin[1] - '1'; 24 beginN.step = 0; 25 endN.x = end[0] - 'a'; 26 endN.y = end[1] - '1'; 27 //push begin Node into the queue 28 q.push(beginN); 29 result = 0; 30 int flag = 0; 31 if (beginN.x == endN.x && beginN.y == endN.y) 32 flag = 1; 33 while (!q.empty() && flag == 0) 34 { 35 pre = q.front(); 36 /* 37 if (abs(endN.x - pre.x) + abs(endN.y - pre.y) == 1) 38 { 39 result = pre.step + 3; 40 break; 41 } 42 else if (abs(endN.x - pre.x) == 2 || abs(endN.y - pre.y) == 2) 43 { 44 result = pre.step + 2; 45 break; 46 } 47 */ 48 q.pop(); 49 //find the eight dirctory 50 for (i = 0; i < 8; i++) 51 { 52 cur.x = pre.x + dx[i]; 53 cur.y = pre.y + dy[i]; 54 cur.step = pre.step + 1; 55 /* 56 if (cur.x < 0 || cur.x >= 8 || cur.y < 0 || cur.y >= 8 57 || (abs(endN.x - pre.x) + abs(endN.y - pre.y) < abs(endN.x - cur.x) + abs(endN.y - cur.y) 58 && !(abs(endN.x - pre.x) == 1 && abs(endN.y - pre.y) == 1))) 59 break; 60 */ 61 //if the point is out of index, then u can ignore 62 if (cur.x < 0 || cur.x >= 8 || cur.y < 0 || cur.y >= 8) 63 continue; 64 q.push(cur); 65 //if u can find the it, then mark the flag 66 if (cur.x == endN.x && cur.y == endN.y) 67 { 68 flag = 1; 69 result = cur.step; 70 break; 71 } 72 } 73 } 74 printf("To get from %s to %s takes %d knight moves.\n", begin, end, result); 75 } 76 return 0; 77 } |
