半平面交 poj1279

Source Code

Problem: 1279		User: Faker_fan
Memory: 296K		Time: 0MS
Language: C++		Result: Accepted

• Source Code

#include <iostream>
#include <string.h>
#include <queue>
#include <stdio.h>
#include <algorithm>
#include <math.h>
typedef long long ll;
using namespace std;
const int maxn=1e4+10;
const int mod=998244353;
using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    double operator ==(const Point &b)const
    {
        return x==b.x&&y==b.y;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
};
double xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2
{
    return (p1-p0)^(p2-p0);
}
double xmult(Point p0,Point p1,Point p2,Point p3) //p0p1 X p2p3
{
    return (p1-p0)^(p3-p2);
}
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
{
    return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0;
}
Point getIntersectPoint(Line a, Line b) {
  double a1 = a.s.y - a.e.y, b1 = a.e.x - a.s.x, c1 = a.s.x * a.e.y - a.e.x * a.s.y;
  double a2 = b.s.y - b.e.y, b2 = b.e.x - b.s.x, c2 = b.s.x * b.e.y - b.e.x * b.s.y;
  return Point((c1*b2-c2*b1)/(a2*b1-a1*b2), (a2*c1-a1*c2)/(a1*b2-a2*b1));
}
double dist(Point a,Point b)
{
    return sqrt( (b - a)*(b - a) );
}


//半平面交s
int neihe_num;
Point dad_anticlock[maxn],neihe[maxn];//输入弄好就行了
Line dad_half_line[maxn],que[maxn];
    //得到极角角度
double getAngle(Point a) {
  return atan2(a.y, a.x);
}
    

//得到极角角度
double getAngle(Line a) {
  return atan2(a.e.y - a.s.y, a.e.x - a.s.x);
}
bool onRight(Line a, Line b, Line c) {
  Point o = getIntersectPoint(b, c);
  if (((a.e - a.s) ^ (o - a.s)) < 0) return true;
  return false;
}
bool cmp_half(Line L1,Line L2)
{
    Point va = L1.e - L1.s, vb = L2.e - L2.s;
    double A =  getAngle(va), B = getAngle(vb);
    if (fabs((double)(A - B)) < eps) return ((va) ^ (L2.e - L1.s)) >= 0;
    return A < B;
}
bool Is_half(int num)
{
    for(int i=0;i<num;i++)
    {
        dad_half_line[i].s=dad_anticlock[i];
        dad_half_line[i].e=dad_anticlock[(i+1)%num];
    }
    sort(dad_half_line, dad_half_line+num, cmp_half);
    
    int head = 0, tail = 0, cnt = 0;//模拟双端队列
    //去重，极角相同时取最后一个。
    for (int i = 0; i < num - 1; i++) {
        if (fabs(getAngle(dad_half_line[i]) - getAngle(dad_half_line[i + 1])) < eps) {
            continue;
        }
        dad_half_line[cnt++] = dad_half_line[i];
    }
    dad_half_line[cnt++] = dad_half_line[num - 1];
    
    
    for (int i = 0; i < cnt; i++) {
        //判断新加入直线产生的影响
        while(tail - head > 1 && onRight(dad_half_line[i], que[tail - 1], que[tail - 2])) tail--;
        while(tail - head > 1 && onRight(dad_half_line[i], que[head], que[head + 1])) head++;
        que[tail++] = dad_half_line[i];
      }
      //最后判断最先加入的直线和最后的直线的影响
      while(tail - head > 1 && onRight(que[head], que[tail - 1], que[tail - 2])) tail--;
      while(tail - head > 1 && onRight(que[tail - 1], que[head], que[head + 1])) head++;
      if (tail - head < 3) return false;
    neihe[0]=getIntersectPoint(que[head],que[tail-1]);
    for(int i=1;i<tail-head;i++)
        neihe[i]=getIntersectPoint(que[i+head],que[(i+head-1)]);
    neihe_num=tail-head;
      return true;//最后拿到的是一些构成内核的线段
}

//半平面交e

//多边形面积s
double mianji_duobianxing(Point *a,int num)
{
    double res=0;
    for(int i=0;i<num;i++)
    {
        res+=a[i]^a[(i+1)%num];
    }
    res/=2;
    return abs(res);
}
//多边形面积e
int main()
{
    int n,T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>dad_anticlock[n-i-1].x>>dad_anticlock[n-i-1].y;
            
        }
        
        if(Is_half(n))
        {
            double res=0;
            res=mianji_duobianxing(neihe, neihe_num);
            printf("%.2lf\n",res);
        }
        else
        {
            cout<<"0.00"<<endl;
        }
    }
}