程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> Codeforces gym 100517（二分，同方向判斷）

Codeforces gym 100517（二分，同方向判斷）

編輯：關於C++

題意：給了一個凸包，按順時針順序給點，點數不超過10萬，再給了兩個不同點，點嚴格在凸包內，凸包保證沒有三點共線，問凸包上有多少對點（pi, pj），滿足pi和pj的線段與兩個點的線段嚴格相交，線段間嚴格相交意思是交點不在端點。
鏈接：http://codeforces.com/gym/100517 (K題)

解法：設凸包有n個點，將凸包點集p擴大一倍，變為2n個點。枚舉前n個點，每次枚舉到 i ，在[i+1, i+n-1]內進行二分，找到兩個點p1,p2，滿足p1和p2是”最靠近” 那兩點線段的點。統計下中間個數即可。二分過程中需要進行一種同方向判斷，即判斷該直線是否在某一個線段”左邊”或”右邊”，用叉積判斷即可。

代碼

//Hello. I'm Peter.
//#pragma comment(linker, /STACK:102400000,102400000)
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const double eps = 1e-9, pi = acos(-1.0);
inline int sgn(double x){
    if(fabs(x) < eps) return 0;
    else return x > 0? 1 : -1;
}
struct Point{
    double x, y;
    Point(){};
    Point(double x1, double y1){x = x1, y = y1;}
};
typedef Point Vector;
Vector operator + (const Vector a, const Vector b){return Vector(a.x + b.x, a.y + b.y);}
Vector operator - (const Vector a, const Vector b){return Vector(a.x - b.x, a.y - b.y);}
double operator * (const Vector a, const Vector b){return a.x * b.x + a.y * b.y;}
double operator % (const Vector a, const Vector b){return a.x * b.y - a.y * b.x;}
Vector operator * (const Vector a, const double b){return Vector(a.x * b, a.y * b);}
Vector operator * (const double b, const Vector a){return Vector(a.x * b, a.y * b);}
Vector operator / (const Vector a, const double b){return Vector(a.x / b, a.y / b);}
bool operator == (const Point a, const Point b){return sgn(a.x - b.x)==0 && sgn(a.y - b.y)==0;}
bool operator || (const Vector a, const Vector b){return sgn(a % b)==0;}
bool operator / (const Vector a, const Vector b){return sgn(a % b)!=0;}
double Length(Vector v){return (double)sqrt((double)(v.x * v.x + v.y * v.y));}
double Dis(Point a, Point b){return Length(a - b);}
Vector Rotate(Vector v, double rad){return Vector(v.x * cos(rad) - v.y * sin(rad), v.x * sin(rad) + v.y * cos(rad));}
double angle(Vector v){return atan2(v.y, v.x);}
double angle(Vector a, Vector b){
    double ans = angle(a) - angle(b);
    while(sgn(ans) < 0) ans += 2*pi; while(sgn(ans) >= 2*pi) ans -= 2*pi;
    return fmin(ans, 2*pi - ans);
}
double Area_Tri(Point p1, Point p2, Point p3){return 0.5 * fabs((p2 - p1) % (p3 - p1));}
double Area_Tri(double a, double b, double c){double p = (a+b+c)/2; return (double)sqrt((double)(p*(p-a)*(p-b)*(p-c)));}

struct Line{
    Point p; Vector v;
    Line(){};
    Line(Point p1, Vector v1){p = p1, v = v1;}
};
Point operator / (const Line a, const Line b){
    double t = ((b.p - a.p) % b.v) / (a.v % b.v);
    return a.p + a.v * t;
}
double Dis(Point p, Line l){return fabs(l.v % (p - l.p)) / Length(l.v);}
double angle(Line a, Line b){double ans = angle(a.v, b.v); return fmin(ans, pi - ans);}

struct Seg{
    Point p1, p2;
    Seg(){};
    Seg(Point p11, Point p22){p1 = p11, p2 = p22;}
};
bool operator / (const Seg a, const Seg b){//need change
    return sgn((a.p2 - a.p1) % (b.p1 - a.p1)) * sgn((a.p2 - a.p1) % (b.p2 - a.p1)) <= 0 &&
    sgn((b.p2 - b.p1) % (a.p1 - b.p1)) * sgn((b.p2 - b.p1) % (a.p2 - b.p1)) <= 0 ;
}
bool operator / (const Line a, const Seg b){
    return sgn(a.v % (b.p1 - a.p)) * sgn(a.v % (b.p2 - a.p)) <= 0;
}
bool PointOnSeg(Point p, Seg s){
    if((s.p1 - p) / (s.p2 - p)) return false;
    else if(sgn((s.p1 - p) * (s.p2 - p)) > 0) return false;
    else return true;
}

#define N 200010
int n;
Point p[N];
Point p1 , p2;
Point readPoi(){
    int x, y;
    x = read(), y = read();
    return Point(x, y);
}
void kuoda(){
    for(int i = n + 1; i <= n + n; i++){
        p[i] = p[i - n];
    }
}

void solve(){
    ll ans = 0;
    for(int i = 1; i <= n; i++){
        int l ,r , mid;
        l = i + 1, r = i + n - 1;
        Vector v1 = p1 - p[i], v2 = p2 - p[i];
        while(l < r){
            mid = (l + r) >> 1;
            Vector v = p[mid] - p[i];
            if(sgn(v % v1) <= 0 && sgn(v % v2) <= 0) r = mid;
            else l = mid + 1;
        }
        int tp1 = l;

        l = i + 1, r = i + n - 1;
        while(l < r){
            mid = (l + r) >> 1;
            mid++;
            Vector v = p[mid] - p[i];
            if(sgn(v % v1) >= 0 && sgn(v % v2) >= 0) l = mid;
            else r = mid - 1;
        }
        int tp2 = l;

        if(tp1 != tp2) ans += tp1 - tp2 - 1;
    }
    ans >>= 1;
    cout< 0){
        for(int i = 1; i <= n; i++) p[i] = readPoi();
        p1 = readPoi();
        p2 = readPoi();

        kuoda();

        solve();
    }
    return 0;
}