Polygons

/*

题意：判断一个顺时针凸包A是否包含(不包括边界)简单多边形B

分析：判断B的每个点是否在A内; 因为n是1e5，m是1e4,如果判断算法在log(m)数量级左右, O(n * Olog(m) )可以过

*/

#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdio>#include <cstdlib>#include <vector>#include <queue>#include <deque>#define MAX 200005#define INF 1e8#define FOR(i, n) for(int i  = 0; i < n; i++)#define FORB(i, n) for(int i = n - 1; i >= 0; i--)#define MP(i, j) make_pair(i, j)#define MEM(array, val) memset(array, val, sizeof(array))#define pu first#define pv second#define eps 1e-50#define INF 1e50using namespace std;inline int cmpd(double x){ return x > eps ? 1 : ( x < - eps ? - 1 : 0); }inline double sqr(double x){ return x * x;}inline bool zero(double x){return cmpd(x) == 0;}int sgn(double x){ return x > eps ? 1 : (x < -eps? -1 : 0); }struct Point{    double x, y;    Point(){}    Point(const  double &xx, const double &yy):x(xx), y(yy){}    Point operator - (const Point &a)const{        return Point(x - a.x, y - a.y);    }    Point operator + (const Point &a)const{        return Point(x + a.x, y + a.y);    }    Point operator / (const double &w)const {        return Point(x / w, y / w);    }    friend double dot(const Point &a, const Point &b){        return a.x *b.x + a.y * b.y;    }    friend double det(const Point &a, const Point &b){        return a.x * b.y - a.y * b.x;    }    void in(){        scanf("%lf %lf",&x, &y);    }    void out(){        printf("%.2f  %.2f\n", x, y);    }};struct Poly{    vector<Point>p;    Poly(){}    bool isContain(const Point &t){        int n = p.size();        Point g = (p[0] + p[n/3] + p[n * 2 / 3]) / 3.0;        int r = n, l = 0;        while(l + 1 < r){            int mid = (l + r)>> 1;            int k =  sgn(det(p[l] - g, p[mid] - g) );            int resl =  sgn( det(p[l] - g, t - g) );            int resm = sgn(det(p[mid] - g, t - g) );            if(k < 0){                if(resl <= 0 && resm > 0){                    r = mid;                }else{                    l = mid;                }            }else {                if(resl > 0 && resm <= 0){                    l  = mid;                }else {                    r = mid;                }            }        }        r %= n;        int z = sgn( det(p[r] - t, p[l] - t) );//        cout<<z<<endl;        if(z == 1){            return true;        }        return false;    }}A;int n, m;int main(){    cin>>n;    Point tt;    FOR(i, n){        tt.in();        A.p.push_back(tt);//顺时针严格凸包    }    bool isInSide = true;    cin>>m;    FOR(j, m){        tt.in();        if(isInSide == true && !A.isContain(tt)){            isInSide= false;        }    }    if(isInSide){        puts("YES");    }else {        puts("NO");    }    return 0;}/*6 -2 1 0 3 3 3 4 1 3 -2 2 -2 4 0 1 2 2 3 1 1 05 1 2 4 2 3 -3 -2 -2 -2 1 4 0 1 1 2 4 1 2 -15 -1 2 2 3 4 1 3 -2 0 -3 5 1 0 1 1 3 1 5 -1 2 -1*/