【HDOJ】1756 Cupid's Arrow

图论，点在多边形内部的判定。

  1 /* 1756 */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23 
 24 #define sti                set<int>
 25 #define stpii            set<pair<int, int> >
 26 #define mpii            map<int,int>
 27 #define vi                vector<int>
 28 #define pii                pair<int,int>
 29 #define vpii            vector<pair<int,int> >
 30 #define rep(i, a, n)     for (int i=a;i<n;++i)
 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 32 #define clr                clear
 33 #define pb                 push_back
 34 #define mp                 make_pair
 35 #define fir                first
 36 #define sec                second
 37 #define all(x)             (x).begin(),(x).end()
 38 #define SZ(x)             ((int)(x).size())
 39 #define lson            l, mid, rt<<1
 40 #define rson            mid+1, r, rt<<1|1
 41 
 42 const double eps = 1e-10;
 43 const int maxn = 105;
 44 
 45 int dcmp(double x) {
 46     if (fabs(x) < eps)    return 0;
 47     return x<0 ? -1:1;
 48 }
 49 
 50 typedef struct Point {
 51     double x, y;
 52     
 53     Point() {}
 54     
 55     Point(double x_, double y_):
 56         x(x_), y(y_) {}
 57 } Point;
 58 
 59 Point poly[maxn];
 60 int n;
 61 
 62 Point operator-(Point A, Point B) {
 63     return Point(A.x-B.x, A.y-B.y);
 64 }
 65 
 66 double Dot(Point A, Point B) {
 67     return A.x*B.x + A.y*B.y;
 68 }
 69 
 70 double Cross(Point A, Point B) {
 71     return A.x*B.y - A.y*B.x;
 72 }
 73 
 74 bool OnSegment(Point P, Point A, Point B) {
 75     return dcmp(Cross(A-P, B-P))==0 && dcmp(Dot(A-P, B-P))<=0;
 76 }
 77 
 78 bool isPointInPolygon(Point p, Point *poly) {
 79     int wn = 0;
 80     
 81     rep(i, 0, n) {
 82         if (OnSegment(p, poly[i], poly[(i+1)%n]))
 83             return true;
 84         int k = dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i]));
 85         int d1 = dcmp(poly[i].y - p.y);
 86         int d2 = dcmp(poly[(i+1)%n].y - p.y);
 87         
 88         if (k>0 && d1<=0 && d2>0)    ++wn;
 89         if (k<0 && d2<=0 && d1>0)    --wn;
 90     }
 91     
 92     return wn!=0;
 93 }
 94 
 95 int main() {
 96     ios::sync_with_stdio(false);
 97     #ifndef ONLINE_JUDGE
 98         freopen("data.in", "r", stdin);
 99         freopen("data.out", "w", stdout);
100     #endif
101     
102     int m;
103     bool flag;
104     Point p;
105     
106     while (scanf("%d",&n) != EOF) {
107         rep(i, 0, n)
108             scanf("%lf %lf", &poly[i].x, &poly[i].y);
109         scanf("%d", &m);
110         while (m--) {
111             scanf("%lf %lf", &p.x, &p.y);
112             flag = isPointInPolygon(p, poly);
113             puts(flag ? "Yes":"No");
114         }
115     }
116     
117     #ifndef ONLINE_JUDGE
118         printf("time = %d.\n", (int)clock());
119     #endif
120     
121     return 0;
122 }

 

posted on 2015-11-01 11:30  Bombe  阅读(142)  评论(0)  编辑  收藏  举报

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