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。
1 #include <cstdio>
2 #include <cstring>
3 #include <cstdlib>
4 #include <cmath>
5 #include <deque>
6 #include <vector>
7 #include <queue>
8 #include <iostream>
9 #include <algorithm>
10 #include <map>
11 #include <set>
12 #include <ctime>
13 #include <iomanip>
14 using namespace std;
15 typedef long long LL;
16 typedef double DB;
17 #define For(i, s, t) for(int i = (s); i <= (t); i++)
18 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
19 #define Rep(i, t) for(int i = (0); i < (t); i++)
20 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
21 #define rep(i, x, t) for(int i = (x); i < (t); i++)
22 #define MIT (2147483647)
23 #define INF (1000000001)
24 #define MLL (1000000000000000001LL)
25 #define sz(x) ((int) (x).size())
26 #define clr(x, y) memset(x, y, sizeof(x))
27 #define puf push_front
28 #define pub push_back
29 #define pof pop_front
30 #define pob pop_back
31 #define ft first
32 #define sd second
33 #define mk make_pair
34 inline void SetIO(string Name)
35 {
36 string Input = Name+".in",
37 Output = Name+".out";
38 freopen(Input.c_str(), "r", stdin),
39 freopen(Output.c_str(), "w", stdout);
40 }
41
42 inline int Getint()
43 {
44 int Ret = 0;
45 char Ch = ‘ ‘;
46 bool Flag = 0;
47 while(!(Ch >= ‘0‘ && Ch <= ‘9‘))
48 {
49 if(Ch == ‘-‘) Flag ^= 1;
50 Ch = getchar();
51 }
52 while(Ch >= ‘0‘ && Ch <= ‘9‘)
53 {
54 Ret = Ret * 10 + Ch - ‘0‘;
55 Ch = getchar();
56 }
57 return Flag ? -Ret : Ret;
58 }
59
60 const int N = 200010;
61 struct Point
62 {
63 DB x, y;
64 inline void Read()
65 {
66 scanf("%lf%lf", &x, &y);
67 }
68 } Arr[N];
69 int n;
70
71 inline void Input()
72 {
73 scanf("%d", &n);
74 For(i, 1, n) Arr[i].Read();
75 }
76
77 inline DB Det(const Point &A, const Point &O, const Point &B)
78 {
79 DB X1 = A.x - O.x, X2 = B.x - O.x;
80 DB Y1 = A.y - O.y, Y2 = B.y - O.y;
81 return X1 * Y2 - X2 * Y1;
82 }
83
84 inline void Solve()
85 {
86 int p = 1;
87 For(i, 2, n)
88 if(Arr[i].x > Arr[p].x) p = i;
89 Arr[0] = Arr[n], Arr[n + 1] = Arr[1];
90 if(Det(Arr[p - 1], Arr[p], Arr[p + 1]) >= 0.0) puts("cw");
91 else puts("ccw");
92 }
93
94 int main()
95 {
96 #ifndef ONLINE_JUDGE
97 SetIO("E");
98 #endif
99 Input();
100 Solve();
101 return 0;
102 }