ZOJ 1091 Knight Moves(BFS)

Knight Moves

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 Specification

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 Specification

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.

 

  1 //Problem Name: Knight Moves
  2 //Source: ZOJ 1091
  3 //Author: jinjin18
  4 //Main idea:BFS
  5 //Language: C++
  6 //Point: init;deal with level in BFS--tail;input--gets rather than scanf;
  7 //-upline input--sscanf;BFS model;
  8 //======================================================================
  9 
 10 #include<stdio.h>
 11 #define MAXSIZE 100000
 12 #define INF 10000000
 13 
 14 typedef struct node{
 15     char x;
 16     int y;
 17 }Point;
 18 int visited[9][9];
 19 void init(){                        //初始化,需要二重循环,可用memset函数
 20     for(int i = 0; i < 9 ;i++){
 21         for(int j = 0; j <9; j++){
 22             visited[i][j] = 0;
 23         }
 24     }
 25 }
 26 int BFS(char ax,int ay,char bx,int by){
 27     int res = 0;
 28     int tail = 1;
 29     visited[ax-'a'+1][ay] = 1;
 30     Point Q[MAXSIZE] = {0};
 31     int pre = 0;
 32     int last = 1;
 33     Q[0].x = ax;
 34     Q[0].y = ay;
 35     while(pre < last){
 36         //printf("OK");
 37         char thisx = Q[pre].x;
 38         int thisy = Q[pre].y;
 39         pre++;
 40         //printf("%d %c %d\n",res,thisx,thisy);
 41         if(thisx == bx&&thisy == by){       //结束循环
 42             return res;
 43         }
 44         if(thisx + 2 <= 'h'&&thisy + 1 <= 8&&visited[thisx+2-'a'+1][thisy+1]==0){
 45             visited[thisx+2-'a'+1][thisy+1]=1;
 46             Q[last].x = thisx+2;
 47             Q[last].y = thisy+1;
 48             last++;
 49         }
 50         if(thisx + 2 <= 'h'&&thisy - 1 >= 1&&visited[thisx+2-'a'+1][thisy-1]==0){
 51             visited[thisx+2-'a'+1][thisy-1]=1;
 52             Q[last].x = thisx+2;
 53             Q[last].y = thisy-1;
 54             last++;
 55         }
 56         if(thisx + 1 <= 'h'&&thisy + 2 <= 8&&visited[thisx+1-'a'+1][thisy+2]==0){
 57             visited[thisx+1-'a'+1][thisy+2]=1;
 58             Q[last].x = thisx+1;
 59             Q[last].y = thisy+2;
 60             last++;
 61         }
 62         if(thisx + 1 <= 'h'&&thisy - 2 >= 1&&visited[thisx+1-'a'+1][thisy-2]==0){
 63             visited[thisx+1-'a'+1][thisy-2]=1;
 64             Q[last].x = thisx+1;
 65             Q[last].y = thisy-2;
 66             last++;
 67         }
 68         if(thisx - 2 >= 'a'&&thisy - 1 >= 1&&visited[thisx-2-'a'+1][thisy-1]==0){
 69             visited[thisx-2-'a'+1][thisy-1]=1;
 70             Q[last].x = thisx-2;
 71             Q[last].y = thisy-1;
 72             last++;
 73         }
 74         if(thisx - 2 >= 'a'&&thisy + 1 <= 8&&visited[thisx-2-'a'+1][thisy+1]==0){
 75             visited[thisx-2-'a'+1][thisy+1]=1;
 76             Q[last].x = thisx-2;
 77             Q[last].y = thisy+1;
 78             last++;
 79         }
 80 
 81         if(thisx - 1 >= 'a'&&thisy - 2 >=1&&visited[thisx-1-'a'+1][thisy-2]==0){
 82             visited[thisx-1-'a'+1][thisy-2]=1;
 83             Q[last].x = thisx-1;
 84             Q[last].y = thisy-2;
 85             last++;
 86         }
 87         if(thisx - 1 >= 'a'&&thisy + 2 <= 8&&visited[thisx-1-'a'+1][thisy+2]==0){
 88             visited[thisx-1-'a'+1][thisy+2]=1;
 89             Q[last].x = thisx-1;
 90             Q[last].y = thisy+2;
 91             last++;
 92         }
 93         if(tail == pre){          //更新tail
 94             tail = last;
 95             res++;
 96         }
 97 
 98     }
 99     return INF;
100 
101 }
102 
103 int main(){
104     int ay,by;
105     char ax,bx;
106     char S[10];
107     while(gets(S)){      //%s与%c不能用,思考为何?
108         sscanf(S,"%c%d %c%d",&ax,&ay,&bx,&by);
109         init();
110         int res = BFS(ax,ay,bx,by);
111         printf("To get from %c%d to %c%d takes %d knight moves.\n",ax,ay,bx,by,res);
112     }
113     return 0;
114 
115 }

看到一篇写的比较好的博文:https://blog.csdn.net/z8110/article/details/49492479

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转载自www.cnblogs.com/jinjin-2018/p/9002132.html