poj 2243 Knight Moves

// 题意:图的大小8*8，可以向 8 个方向移动,求起点到终点的最短移动距离 
#include <iostream>        // BFS
#include <string>
using namespace std;
const int L=8;
int q[L*L],vis[L][L],dist[L][L];
const int dx[]={-1,-2,-2,-1,1,2,2,1};    
const int dy[]={-2,-1,1,2,2,1,-1,-2};
int sx,sy,tx,ty;
void bfs()
{
    vis[sx][sy]=1;
    dist[sx][sy]=0;
    q[0]=sx*L+sy;
    int front=0,rear=1;
    while(front<rear)
    {
        int z=q[front];
        int x=z/L,y=z%L;
        if(x==tx&&y==ty)
        {
            printf("To get from %c%d to %c%d takes %d knight moves.\n",sx+'a',sy+1,tx+'a',ty+1,dist[tx][ty]);
            break;
        }
        for(int d=0;d<8;++d)
        {
            int newx=x+dx[d];
            int newy=y+dy[d];
            int newz=newx*L+newy;
            if(newx>=0&&newx<L&&newy>=0&&newy<L&& !vis[newx][newy] )
            {
                vis[newx][newy]=1;
                q[rear]=newz;
                dist[newx][newy]=dist[x][y]+1;
                rear++;
            }
        }
        front++;
    }
}
int main()
{
    char ch[10];
    while(gets(ch))
    {
        sx=ch[0]-'a';
        sy=ch[1]-'1';
        tx=ch[3]-'a';
        ty=ch[4]-'1';
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}

 

 

 

 

#include <iostream>        // 双向BFS
#include <string>
using namespace std;
const int L=8;
int q1[L*L],q2[L*L],vis[L][L],dist1[L][L],dist2[L][L];    
const int dx[]={-1,-2,-2,-1,1,2,2,1};    
const int dy[]={-2,-1,1,2,2,1,-1,-2};
int sx,sy,tx,ty;
void bfs()
{
    if(sx==tx&&sy==ty)
    {
        printf("To get from %c%d to %c%d takes 0 knight moves.\n",sx+'a',sy+1,tx+'a',ty+1);
        return ;
    }
    vis[sx][sy]+=1;
    dist1[sx][sy]=0;
    q1[0]=sx*L+sy;    
    vis[tx][ty]+=2;
    dist2[tx][ty]=0;
    q2[0]=tx*L+ty;
    int front1=0,rear1=1,front2=0,rear2=1,ok=0;
    while(front1<rear1||front2<rear2)
    {

        //正向搜索
        if(front1<rear1)
        {
            int z=q1[front1];
            int x=z/L,y=z%L;
            for(int d=0;d<8;++d)
            {
                int newx=x+dx[d];
                int newy=y+dy[d];
                int newz=newx*L+newy;
                if(newx>=0&&newx<L&&newy>=0&&newy<L)    
                {
                    if( vis[newx][newy] == 0 )
                    {
                        vis[newx][newy] = 1 ;    // 正向搜索标记为 1 
                        q1[rear1]=newz;
                        dist1[newx][newy]=dist1[x][y]+1;
                        rear1++;
                    }
                    else if( vis[newx][newy] == 2 )        // 说明正向和反向搜索发生相遇
                    {
                        dist1[newx][newy]=dist1[x][y]+1;
                        printf("To get from %c%d to %c%d takes %d knight moves.\n",sx+'a',sy+1,tx+'a',ty+1,dist1[newx][newy]+dist2[newx][newy]);
                        ok=1;
                        break;
                    }
                }
            }
            front1++;

            if(ok)
                break;
        }

        //反向搜索
        if(front2<rear2)
        {
            int z=q2[front2];
            int x=z/L,y=z%L;
            for(int d=0;d<8;++d)
            {
                int newx=x+dx[d];
                int newy=y+dy[d];
                int newz=newx*L+newy;
                if(newx>=0&&newx<L&&newy>=0&&newy<L)    
                {
                    if( vis[newx][newy] == 0 )
                    {
                        vis[newx][newy] = 2 ;    // 反向搜索标记为 2 
                        q2[rear2]=newz;
                        dist2[newx][newy]=dist2[x][y]+1;
                        rear2++;
                    }
                    else if( vis[newx][newy] == 1 )
                    {
                        dist2[newx][newy]=dist2[x][y]+1;
                        printf("To get from %c%d to %c%d takes %d knight moves.\n",sx+'a',sy+1,tx+'a',ty+1,dist1[newx][newy]+dist2[newx][newy]);
                        ok=1;
                        break;
                    }
                }
            }
            front2++;

            if(ok)
                break;
        }

    }
}
int main()
{
    char ch[10];
    while(gets(ch))
    {
        sx=ch[0]-'a';
        sy=ch[1]-'1';
        tx=ch[3]-'a';
        ty=ch[4]-'1';
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}