POJ1835 宇航员模拟

　　模拟，很无聊的题目-_-||，把转换方向映射好就可以了，我是对宇航员保存三个方向的向量，那么每次变化方向的时候就方便了。
  1 //STATUS:C++_AC_188MS_172KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //define
 25 #define pii pair<int,int>
 26 #define mem(a,b) memset(a,b,sizeof(a))
 27 #define lson l,mid,rt<<1
 28 #define rson mid+1,r,rt<<1|1
 29 #define PI acos(-1.0)
 30 //typedef
 31 typedef __int64 LL;
 32 typedef unsigned __int64 ULL;
 33 //const
 34 const int N=1010;
 35 const int INF=0x3f3f3f3f;
 36 const int MOD=256,STA=8000010;
 37 const LL LNF=1LL<<60;
 38 const double EPS=1e-8;
 39 const double OO=1e15;
 40 const int dx[4]={-1,0,1,0};
 41 const int dy[4]={0,1,0,-1};
 42 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 43 //Daily Use ...
 44 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 45 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 46 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 47 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 48 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 49 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 50 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 51 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 52 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 53 //End
 54 
 55 struct Node{
 56     int x,y,z;
 57 }r1[3],r2[3];
 58 int T,n;
 59 
 60 void re(Node &a,Node b)
 61 {
 62     a.x=-b.x;
 63     a.y=-b.y;
 64     a.z=-b.z;
 65 }
 66 
 67 int main()
 68 {
 69  //   freopen("in.txt","r",stdin);
 70     int i,j,d,x,y,z;
 71     char s[20];
 72     Node *p,*q;
 73     scanf("%d",&T);
 74     while(T--)
 75     {
 76         x=y=z=0;
 77         r1[0].x=1,r1[0].y=0,r1[0].z=0;
 78         r1[1].x=0,r1[1].y=1,r1[1].z=0;
 79         r1[2].x=0,r1[2].y=0,r1[2].z=1;
 80         scanf("%d",&n);
 81         while(n--){
 82             scanf("%s%d",s,&d);
 83             if(s[0]=='f'){
 84                 r2[0]=r1[0],r2[1]=r1[1],r2[2]=r1[2];
 85             }
 86             else if(s[0]=='b'){
 87                 re(r2[0],r1[0]);
 88                 re(r2[1],r1[1]);
 89                 r2[2]=r1[2];
 90             }
 91             else if(s[0]=='l'){
 92                 re(r2[0],r1[1]);
 93                 r2[1]=r1[0];
 94                 r2[2]=r1[2];
 95             }
 96             else if(s[0]=='r'){
 97                 r2[0]=r1[1];
 98                 re(r2[1],r1[0]);
 99                 r2[2]=r1[2];
100             }
101             else if(s[0]=='u'){
102                 r2[0]=r1[2];
103                 r2[1]=r1[1];
104                 re(r2[2],r1[0]);
105             }
106             else {
107                 re(r2[0],r1[2]);
108                 r2[1]=r1[1];
109                 r2[2]=r1[0];
110             }
111 
112             x+=r2[0].x*d;
113             y+=r2[0].y*d;
114             z+=r2[0].z*d;
115             swap(r2[0],r1[0]);
116             swap(r2[1],r1[1]);
117             swap(r2[2],r1[2]);
118         }
119 
120         int ans;
121         if(r1[0].x==1)ans=0;
122         else if(r1[0].y==1)ans=1;
123         else if(r1[0].z==1)ans=2;
124         else if(r1[0].x==-1)ans=3;
125         else if(r1[0].y==-1)ans=4;
126         else ans=5;
127 
128         printf("%d %d %d %d\n",x,y,z,ans);
129     }
130     return 0;
131 }
POJ1835 宇航员 模拟

POJ1835 宇航员模拟
