POJ 2243 Knight Moves

Knight Moves

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13823		Accepted: 7719

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 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.

Source

Ulm Local 1996

#include <iostream>
#include <queue>
using namespace std;
struct knight {
    int x,y,step;
    int g,h,f;
    bool operator < (const knight & k) const {     
        return f > k.f;
    }
} k;
bool visited[8][8];                               
int x1,y1,x2,y2,ans;                               //起点(x1,y1),终点(x2,y2),最少移动次数ans
int 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ceil
                s.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.
",line[0],line[1],line[3],line[4],ans);
    }
    return 0;
}

听人说A*算法很厉害（可以大幅度提高搜索的效率，快到你都想不到），但一听说要有一步预判，我就想就凭计算机那智商~~~~~-_-  -_-  -_-然而还是在今晚自学了一下A* 搜索，感觉还是不很不靠谱，时间复杂度：玄学（学长们都这么说~~QWQ）