杭电 HOJ 1372 Knight Moves 解题报告

    马走棋盘，搜索。刚开始想打表，发现棋子在边界的时候很难处理，比如a1到b2的话2步是不行的。

    用队列，时效差一点点，46MS，代码如下：

#include<iostream>

#include<deque>

using namespace std;



struct Point

{

    char x,y,num;

};



int main()

{

    int a,b,i,j;

    Point x,y;

    deque<Point> q;

    char str[2],str2[2];

    char visit[8][8];

    while(cin>>str>>str2)

    {

        memset(visit,0,sizeof(visit));

        x.x=str[0]-'a';

        x.y=str[1]-'1';

        x.num=0;

        q.clear();

        q.push_back(x);

        visit[x.x][x.y]=1;

        a=str2[0]-'a';

        b=str2[1]-'1';



        while(!(x.x==a&&x.y==b))

        {

            y.num=x.num+1;

            if(x.x>=2 && x.y>=1 && !visit[y.x=x.x-2][y.y=x.y-1])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x<6 && x.y>=1 && !visit[y.x=x.x+2][y.y=x.y-1])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x<6 && x.y<7 && !visit[y.x=x.x+2][y.y=x.y+1])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x>=2 && x.y<7 && !visit[y.x=x.x-2][y.y=x.y+1])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x>=1 && x.y>=2 && !visit[y.x=x.x-1][y.y=x.y-2])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x>=1 && x.y<6 && !visit[y.x=x.x-1][y.y=x.y+2])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x<7 && x.y<6 && !visit[y.x=x.x+1][y.y=x.y+2])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }

            if(x.x<7 && x.y>=2 && !visit[y.x=x.x+1][y.y=x.y-2])

            {

                q.push_back(y);

                visit[y.x][y.y]=1;

            }



            q.pop_front();

            x=q.front();

        }

        cout<<"To get from "<<str<<" to "<<str2<<" takes "<<(int)x.num<<" knight moves."<<endl;

    }

}

    忽然发现自己曾经过了这题，代码是网上找的，也贴上吧~思路是一样的，循环部分比我的简洁。

#include<iostream>  

#include<cstdio>  

#include<cmath>  

#include<cstring>  

#include<cstdlib>  

#include<cctype>  

#include<iomanip>  

#include<string>  

#include<algorithm>  

#include<ctime>  

#include<stack>  

#include<queue>  

#include<vector>  

class Cordinate  

{public:int x,y;};  

using namespace std;  

int dir[8][2]={1,2,1,-2,2,1,2,-1,-1,2,-1,-2,-2,1,-2,-1},chess[10][10];  

bool legal(Cordinate po)  

{if(po.x>=1&&po.x<=8&&po.y>=1&&po.y<=8)return true;return false;}  

int main()  

{



    char a[3],b[3];Cordinate po,pos;int i;  

    while(cin>>a>>b)  

    {  

        queue<Cordinate>q;  

        memset(chess,-1,sizeof(chess));  

        pos.x=a[0]-'a'+1;pos.y=a[1]-'0';  

        chess[pos.x][pos.y]=0;  

        q.push(pos);  

        while(!q.empty())  

        {  

            po=q.front();q.pop();  

            for(i=0;i<8;i++)  

            {  

                pos.x=dir[i][0]+po.x;pos.y=dir[i][1]+po.y;  

                if(legal(pos)&&chess[pos.x][pos.y]==-1) {q.push(pos);chess[pos.x][pos.y]=chess[po.x][po.y]+1;}  

            }  

        }  

        cout<<"To get from "<<a<<" to "<<b<<" takes "<<chess[b[0]-'a'+1][b[1]-'0']<<" knight moves."<<endl;  

    }  

    return 0;  

}