ural 1246. Tethered Dog

1246. Tethered Dog

Time limit: 1.0 second
Memory limit: 64 MB

A dog is tethered to a pole with a rope. The pole is located inside a fenced polygon (not necessarily convex) with nonzero area. The fence has no self-crosses. The Olympian runs along the fence bypassing the vertices of the polygon in a certain order which is not broken during the jog. A dog pursues him inside the fenced territory and barks. Your program is to determine how (clockwise or counter-clockwise) the rope will wind after several rounds of the Olympian's jog.

Input

The first input line contains a number N that is the number of the polygon vertices. It’s known that 3 ≤ N ≤ 200000. The next N lines consist of the vertices plane coordinates, given in an order of Olympian’s dog. The coordinates are a pair of integers separated with a space. The absolute value of each coordinate doesn’t exceed 50000.

Output

You are to output "cw", if the rope is winded in a clockwise order and "ccw" otherwise.

Sample

input	output
4 0 0 0 1 1 1 1 0	cw

Problem Author: Evgeny Kobzev
Problem Source: Ural State University Personal Programming Contest, March 1, 2003 

Tags: geometry  (hide tags for unsolved problems)

Difficulty: 364 

 

题意：问一个多边形的给出顺序是不是顺时针。

分析：叉积判断方向。用最左或最优的点作为基准点。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 #define Rep(i, t) for(int i = (0); i < (t); i++)
 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair
 inline void SetIO(string Name)
 {
     string Input = Name+".in",
     Output = Name+".out";
     freopen(Input.c_str(), "r", stdin),
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 struct Point
 {
     DB x, y;
     inline void Read()
     {
         scanf("%lf%lf", &x, &y);
     }
 } Arr[N];
 int n;

 inline void Input()
 {
     scanf("%d", &n);
     For(i, , n) Arr[i].Read();
 }

 inline DB Det(const Point &A, const Point &O, const Point &B)
 {
     DB X1 = A.x - O.x, X2 = B.x - O.x;
     DB Y1 = A.y - O.y, Y2 = B.y - O.y;
     return X1 * Y2 - X2 * Y1;
 }

 inline void Solve()
 {
     int p = ;
     For(i, , n)
         if(Arr[i].x > Arr[p].x) p = i;
     Arr[] = Arr[n], Arr[n + ] = Arr[];
     if(Det(Arr[p - ], Arr[p], Arr[p + ]) >= 0.0) puts("cw");
     else puts("ccw");
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
     SetIO("E");
     #endif
     Input();
     Solve();
     return ;
 }