HDU 1372 Knight moves

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14208    Accepted Submission(s): 8306

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.

 

题意：在一个 8 × 8 的网格中，给你两个点，按马行日的方法走，求走的最短步数，注意下输出的格式，别只输出了结果没有输出那句话。。。

注意：马走日，有八个方向！

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
 9 char s1[3],s2[3];
10 int sx,sy,ex,ey;
11 bool mark[10][10];
12 struct node
13 {
14     int x,y;
15     int step;
16 };
17 
18 void bfs()
19 {
20     queue<node> q;
21     struct node cu,ne;
22     cu.x = sx;
23     cu.y = sy;
24     cu.step =0;
25     mark[cu.x][cu.y] = 1; //别忘了标记，虽然本题不标记也可以；
26     q.push(cu);
27     while(!q.empty())
28     {
29         cu = q.front();
30         q.pop();
31         if(cu.x == ex && cu.y == ey)
32         {
33             printf("To get from %s to %s takes %d knight moves.\n",s1,s2,cu.step);
34             return ;
35         }
36         for(int i=0;i<8;i++)
37         {
38             ne.x = cu.x + go[i][0];
39             ne.y = cu.y + go[i][1];
40             if(ne.x>=0&&ne.x<8&&ne.y>=0&&ne.y<8&&!mark[ne.x][ne.y]) //注意这个范围是0--7的；
41             {
42                 ne.step = cu.step + 1;
43                 mark[ne.x][ne.y] = 1;
44                 q.push(ne);
45             }
46         }
47     }
48 }
49 int main()
50 {
51     //也可以用字母加数字的方式输入。eg while(scanf("%c%d %c%d",&c1,&d1,&c2,&d2),但别忘了输出也是字母加数字
52     while(scanf("%s%s",s1,s2)!=EOF) //这两个%s之间的空格有没有都可以
53     {
54         //a--h 和 1--8 数量是一样的，所以，a--h减'a'相当于0--7，所以，数字别忘了减 1 ；
55         sx = s1[0] - 'a';
56         sy = s1[1] - '1';
57         ex = s2[0] - 'a';
58         ey = s2[1] - '1';
59         memset(mark,0,sizeof(mark));
60         bfs();
61         memset(s1,0,sizeof(s1));
62         memset(s2,0,sizeof(s2));
63     }
64     return 0;
65 }

Ac代码

 

转载于:https://www.cnblogs.com/cypblogs/p/10023261.html