POJ 2243 A*算法

如果朋友你对A*算法不理解，请读原著，写的真好。

原著：http://www.cppblog.com/mythit/archive/2009/04/19/80492.aspx

 

原题：http://acm.pku.edu.cn/JudgeOnline/problem?id=2243
problem statement

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.
题目的意思大概是说：在国际象棋的棋盘上，一匹马共有8个可能的跳跃方向，求从起点到目标点之间的最少跳跃次数。

代码：

#include <iostream>#include <queue>using namespace std;struct knight{int x,y,step;int g,h,f;bool operator < (const knight & k) const{      //重载比较运算符return f > k.f;}}k;bool visited[8][8];                                //已访问标记(关闭列表)int x1,y1,x2,y2,ans;                               //起点(x1,y1),终点(x2,y2),最少移动次数ansint dirs[8][2]={{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};//8个移动方向priority_queue<knight> que;                        //最小优先级队列(开启列表)bool in(const knight & a){                         //判断knight是否在棋盘内if(a.x<0 || a.y<0 || a.x>=8 || a.y>=8)return false;return true;}int Heuristic(const knight &a){                    //manhattan估价函数return (abs(a.x-x2)+abs(a.y-y2))*10;}void Astar(){                                      //A*算法knight t,s;while(!que.empty()){t=que.top(),que.pop(),visited[t.x][t.y]=true;if(t.x==x2 && t.y==y2){ans=t.step;break;}for(int i=0;i<8;i++){s.x=t.x+dirs[i][0],s.y=t.y+dirs[i][1];if(in(s) && !visited[s.x][s.y]){s.g = t.g + 23;                 //23表示根号5乘以10再取其ceils.h = Heuristic(s);s.f = s.g + s.h;s.step = t.step + 1;que.push(s);}}}}int main(){char line[5];while(gets(line)){x1=line[0]-'a',y1=line[1]-'1',x2=line[3]-'a',y2=line[4]-'1';memset(visited,false,sizeof(visited));k.x=x1,k.y=y1,k.g=k.step=0,k.h=Heuristic(k),k.f=k.g+k.h;while(!que.empty()) que.pop();que.push(k);Astar();printf("To get from %c%c to %c%c takes %d knight moves.\n",line[0],line[1],line[3],line[4],ans);    }    return 0;}