BFS之五(双向bfs)

#include <iostream>            //poj 2243 Knight Moves  参照poj 1915
#include <deque>
using namespace std;
int dir[2][8]={{-2,-1,1,2,2,1,-1,-2},{-1,-2,-2,-1,1,2,2,1}};
int visited[10][20],path[10][20];
struct node
{
    int x,y,c;
    bool operator==(const node& other)
    {
        return x==other.x&&y==other.y;
    }
    node(int xx=0,int yy=0,int cc=0)
    {
        c=cc;x=xx;y=yy;
    }
};
deque<node> forword,backword;
int main()
{
    //freopen("F:\\c++.txt", "r", stdin ) ;
    char in[10];
    int x,y,c,tx,ty;
    while(gets(in))
    {
        memset(visited,0,sizeof(visited));
        memset(path,0,sizeof(path));
        node beg(int(in[1]-48),int(in[0]-96)),end(int(in[4]-48),int(in[3]-96));
        if(beg==end)
        {
            printf("To get from %c%c to %c%c takes 0 knight moves.\n",in[0],in[1],in[3],in[4]);
            continue;
        }
        visited[beg.x][beg.y]=1;visited[end.x][end.y]=2;
        forword.clear();backword.clear();
        backword.push_back(end);forword.push_back(beg);
        char tag=0;
        while(!tag&&!forword.empty())
        {
            node nd1=forword.front();
            forword.pop_front();
            x=nd1.x;y=nd1.y;c=nd1.c+1;    
            for(int i=0;i<8;++i)
            {
                tx=x+dir[0][i];
                ty=y+dir[1][i];
                if(tx<1||tx>8||ty<1||ty>8)
                    continue;
                if(visited[tx][ty]==0)
                {
                    path[tx][ty]=c;
                    visited[tx][ty]=1;
                    node t(tx,ty,c);
                    forword.push_back(t);
                }
                else if(visited[tx][ty]==2)
                {
                    printf("To get from %c%c to %c%c takes %d knight moves.\n",in[0],in[1],in[3],in[4],c+path[tx][ty]);
                    tag=1;break;
                }
            }
            if(tag)
                break;

            if(!backword.empty())
            {
                node nd2=backword.front();
                backword.pop_front();
                x=nd2.x;y=nd2.y;c=nd2.c+1;
                for(int i=0;i<8;++i)
                {
                    tx=x+dir[0][i];
                    ty=y+dir[1][i];
                    if(tx<1||tx>8||ty<1||ty>8)
                        continue;
                    if(visited[tx][ty]==0)
                    {
                        path[tx][ty]=c;
                        visited[tx][ty]=2;
                        node t(tx,ty,c);
                        backword.push_back(t);
                    }
                    else if(visited[tx][ty]==1)
                    {
                        printf("To get from %c%c to %c%c takes %d knight moves.\n",in[0],in[1],in[3],in[4],c+path[tx][ty]);
                        tag=1;break;
                    }
                }
            }
        }
    }
    return 0;
}