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  • 【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.
    ", (int)clock());
    119     #endif
    120     
    121     return 0;
    122 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/4927407.html
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