UVA 439 Knight Moves

最新推荐文章于 2021-07-31 22:20:00 发布

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最新推荐文章于 2021-07-31 22:20:00 发布

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本文链接：https://blog.youkuaiyun.com/xf_zhen/article/details/51584356

本文介绍了一个计算国际象棋中马从一个点到另一个点所需最少步数的算法实现。通过使用广度优先搜索（BFS），该算法能够有效地找到最短路径。代码采用C++编写，包括了读取起点和终点坐标、初始化棋盘状态、进行路径计算及输出结果等功能。

题目：给出国际象棋棋盘中的两个点，求马从一个点跳到另一个点的最少步数。

#include <iostream>
#include<queue>
using namespace std;
struct RA{
    int c;
    int  r;
    RA(int a = 0, int b = 0){c = a; r = b;};
    RA(char a,char b){c = int(a - 'a');r = int(b -'1');};
    char ioc(){return char(c+int('a'));};
    char ioi(){return char(r + int('1'));};
    bool friend operator == (RA a, RA b){return (a.c == b.c)&&(a.r == b.r);};
};

const int SIZE = 8;
class Knight{
private:
    RA SA;
    RA SB;
    int board[SIZE][SIZE];
public:
    void initial();
    bool readData();
    void computing();
    void outResult();
};

void Knight::initial() {
for(int i = 0;i < SIZE; i++){
    for(int j = 0; j < SIZE; j++)
        board[i][j]=-2;
}
}
bool Knight::readData(){
char c,r;
cin>>c>>r;

SA = RA(c,r);
cin>>c>>r;

SB = RA(c, r);
return !cin.eof();
}
void Knight::outResult(){
cout<<"To get from "<<SA.ioc()<<SA.ioi();
cout<<" to "<<SB.ioc()<<SB.ioi()<<" takes ";
cout<<board[SB.c][SB.r]<<" knight moves."<<endl;
}
void Knight::computing(){
RA ns,s;
const int dr[8]={2,1,-1,-2,-2,-1,1,2};
const int dc[8]={1,2,2,1,-1,-2,-2,-1};
queue<RA> q;
board[SA.c][SA.r] = 0;
q.push(SA);
while(!q.empty()){
    s = q.front();
    q.pop();

    if(s == SB)
        break;
    else{
        for(int i = 0; i < 8; i++){
            ns = RA(s.c + dc[i], s.r + dr[i]);
            if((0 <= ns.c)&&(ns.c < 8)&&(0 <= ns.r)&&(ns.r < 8))
            {
                if(board[ns.c][ns.r] < 0){
                    board[ns.c][ns.r] = 1 + board[s.c][s.r] ;
                    q.push(ns);
                }
            }
        }
    }
}
}
int main()
{
    Knight ss;
    while(ss.readData()){
        ss.initial();
        ss.computing();
        ss.outResult();

    }
    return 0;
}