紫书 UVA439

最新推荐文章于 2021-03-20 20:07:00 发布

Dilly__dally

最新推荐文章于 2021-03-20 20:07:00 发布

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分类专栏：紫书第六章习题

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本文介绍了一个使用广度优先搜索算法来计算国际象棋中骑士从一个位置到另一个位置所需最少移动次数的C++程序实现。通过定义骑士可能的移动方向，并采用队列进行节点扩展，该程序能够有效地找到最短路径。

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/*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.*/
#include <bits/stdc++.h>

using namespace std;
int m[9][9];
int dir[8][2]= {{-2,1},{-1,2},{1,2},{2,1},{-2,-1},{-1,-2},{1,-2},{2,-1}};
struct node
{
    int x,y;
    node(int a,int b):x(a),y(b) {}
};
int d[9][9];
int vis[9][9];
char a,b;
int a1,b1;
void bfs(node a)
{
    queue<node> q;
    q.push(a);
    //cout<<a.x<<a.y;
    memset(vis,0,sizeof(vis));
    memset(d,0,sizeof(d));
    vis[a.x][a.y]=1;
    while(!q.empty())
    {
        node p=q.front();
        int x0=p.x;
        int y0=p.y;
        if(p.x==b1&&p.y==b-'a'+1)
            return;
        q.pop();
        for(int i=0; i<8; i++)
        {
            int x1=x0+dir[i][0];
            int y1=y0+dir[i][1];
          //  cout<<x1<<" "<<y1<<endl;;
            if(x1>0&&x1<=8&&y1>0&&y1<=8&&!vis[x1][y1])
            {
                node t(x1,y1);
                d[x1][y1]=d[x0][y0]+1;
                //cout<<d[x1][y1]<<" ";
                if(x1==b1&&y1==(b-'a'+1))
                    return ;
                vis[x1][y1]=1;
                q.push(t);
            }
        }
    }

}
int main()
{

    while(scanf("%c%d %c%d",&a,&a1,&b,&b1)!=EOF)
    {
        getchar();//此处 一定要清除上一次输入的'\n'
        bfs(node(a1,a-'a'+1));
        cout<<"To get from "<<a<<a1<<" to "<<b<<b1<<" takes "<<d[b1][b-'a'+1]<<" knight moves."<<endl;
    }
    return 0;
}