[Codeforces 166B] Polygons （点在凸多边形内）

Codeforces - 166B

判断任意多边形 B是否严格在凸多边形 A内部

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点在凸多边形内部试板题
如果 B的所有顶点在 A内，则 B在 A内
由于 A的顶点有 105个，B的顶点有 104 个
所以不能用 (n)的暴力判断
有一个 (logn) 的二分做法
基本原理是用对角线将凸多边形剖成以 cor[0] 为公共顶点的若干个三角形
然后二分出点 p在哪两条对角线中，最后再用这个三角形的第三条边判断是否在三角形内

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <cctype>#include <map>#include <set>#include <queue>#include <bitset>#include <string>#include <complex>using namespace std;typedef pair<int,int> Pii;typedef long long LL;typedef unsigned long long ULL;typedef double DBL;typedef long double LDBL;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define SQR(a) ((a)*(a))#define PCUT puts("\n----------")const DBL eps = 1e-11;int sgn(DBL x){return x>eps?1:(x<-eps?-1:0);}struct Vector{    DBL x,y;    Vector(DBL _x=0, DBL _y=0):x(_x),y(_y){}    Vector operator + (const Vector &rhs) const {return Vector(x+rhs.x, y+rhs.y);}    Vector operator - (const Vector &rhs) const {return Vector(x-rhs.x, y-rhs.y);}    Vector operator * (const DBL &rhs) const {return Vector(x*rhs, y*rhs);}    Vector operator / (const DBL &rhs) const {return Vector(x/rhs, y/rhs);}    DBL operator * (const Vector &rhs) const {return x*rhs.x + y*rhs.y;}    DBL operator ^ (const Vector &rhs) const {return x*rhs.y - rhs.x*y;}    int read(){return scanf("%lf%lf", &x, &y);}    int pri(){return printf("P:%.2f %.2f\n", x, y);}};typedef Vector Point;struct Polygon{    int siz;    vector<Point> cor;    vector<Vector> tx;    void init(){cor.clear(); tx.clear(); siz=0;}    void read(int _n)    {        siz=_n;        cor.resize(siz);        for(int i=0; i<siz; i++)        {            cor[i].read();            tx.push_back(cor[i]-cor[0]);        }    }};int N;Polygon A;bool PointInPolygon(Point&,Polygon&);int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);    #endif    while(~scanf("%d", &N))    {        A.init(); A.read(N);        scanf("%d", &N);        Point tem;        bool ok=1;        for(int i=0; i<N; i++)        {            tem.read();            ok &= PointInPolygon(tem, A);        }        puts(ok?"YES":"NO");    }    return 0;}bool PointInPolygon(Point &p, Polygon &poly){    if(poly.siz<3) return 0;    Vector tp = p-poly.cor[0];    if(sgn(tp^poly.tx[1]) <= 0) return 0;    if(sgn(tp^poly.tx[poly.siz-1]) >= 0) return 0;    int l=1,r=poly.siz-2,mid;    while(l<r)    {        mid = (l+r+1)>>1;        if(sgn(tp^poly.tx[mid]) >= 0) l = mid;        else r = mid-1;    }    return sgn((p-poly.cor[l])^(poly.cor[l+1]-poly.cor[l])) > 0;}