poj 1815

模拟，标记六个方向的位置为0,1,2。。。，然后记录当前位置六种走法所要朝向的方向，而我们所需要的就是每个位置的这几个方向，很显然，其中的front所得到的方向就是当前所朝向的方向。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;struct node{    int front,back,left,right,up,down;};node b_ch(node op){    node tem;    tem.front=op.back;    tem.back=op.front;    tem.left=op.right;    tem.right=op.left;    tem.up=op.up;    tem.down=op.down;    return tem;}node l_ch(node op){    node tem;    tem.front=op.left;    tem.back=op.right;    tem.left=op.back;    tem.right=op.front;    tem.up=op.up;    tem.down=op.down;    return tem;}node r_ch(node op){    node tem;    tem.front=op.right;    tem.back=op.left;    tem.left=op.front;    tem.right=op.back;    tem.up=op.up;    tem.down=op.down;    return tem;}node u_ch(node op){    node tem;    tem.front=op.up;    tem.back=op.down;    tem.left=op.left;    tem.right=op.right;    tem.up=op.back;    tem.down=op.front;    return tem;}node d_ch(node op){    node tem;    tem.front=op.down;    tem.back=op.up;    tem.left=op.left;    tem.right=op.right;    tem.up=op.front;    tem.down=op.back;    return tem;}int main(){    int  m;    cin>>m;    while(m--)    {        int n;        scanf("%d",&n);        char op[30];        int i,dis;        node td;        td.front=0,td.back=3,td.left=4,td.right=1,td.up=2,td.down=5;        int x=0,y=0,z=0;        for(i=0;i<n;i++)        {            scanf("%s %d",op,&dis);            if(strcmp(op,"forward")==0)            {                td=td;            }            else if(strcmp(op,"back")==0)  {td=b_ch(td);}             else if(strcmp(op,"left")==0) {td=l_ch(td);}            else if(strcmp(op,"right")==0) {td=r_ch(td);}            else if(strcmp(op,"up")==0) {td=u_ch(td);}            else {td=d_ch(td);}            if(td.front==0) {x+=dis;}            else if(td.front==1) {y+=dis;}            else if(td.front==2) {z+=dis;}            else if(td.front==3) {x-=dis;}            else if(td.front==4) {y-=dis;}            else  {z-=dis;}        }        printf("%d %d %d %d\n",x,y,z,td.front );    }    return 0;}