标签:des style blog class code java
1 1.00 2 -1.00 0.00 1.00 0.00 0.00 -1.00 0.00 1.00 0.50 0.50
Case 1: 25.00000%
/* ***********************************************
Author :_rabbit
Created Time :2014/5/4 15:03:55
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-5
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x>0?1:-1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0){
x=_x;y=_y;
}
};
Point operator + (const Point &a,const Point &b){
return Point(a.x+b.x,a.y+b.y);
}
Point operator - (const Point &a,const Point &b){
return Point(a.x-b.x,a.y-b.y);
}
Point operator * (const Point &a,const double &p){
return Point(a.x*p,a.y*p);
}
Point operator / (const Point &a,const double &p){
return Point(a.x/p,a.y/p);
}
bool operator < (const Point &a,const Point &b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
bool operator == (const Point &a,const Point &b){
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Point a,Point b){
return a.x*b.x+a.y*b.y;
}
double Length(Point a){
return sqrt(Dot(a,a));
}
double Angle(Point a,Point b){
return acos(Dot(a,b)/Length(a)/Length(b));
}
double angle(Point a){
return atan2(a.y,a.x);
}
double Cross(Point a,Point b){
return a.x*b.y-a.y*b.x;
}
Point vecunit(Point a){
return a/Length(a);
}
Point Normal(Point a){
return Point(-a.y,a.x)/Length(a);
}
Point Rotate(Point a,double rad){
return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
}
double Area2(Point a,Point b,Point c){
return Length(Cross(b-a,c-a));
}
bool OnSegment(Point p,Point a1,Point a2){
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<=0;
}
struct Line{
Point p,v;
double ang;
Line(){};
Line(Point p,Point v):p(p),v(v){
ang=atan2(v.y,v.x);
}
bool operator < (const Line &L) const {
return ang<L.ang;
}
Point point(double d){
return p+(v*d);
}
};
bool OnLeft(const Line &L,const Point &p){
return Cross(L.v,p-L.p)>=0;
}
Point GetLineIntersection(Point p,Point v,Point q,Point w){
Point u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
Point GetLineIntersection(Line a,Line b){
return GetLineIntersection(a.p,a.v,b.p,b.v);
}
vector<Point> HPI(vector<Line> L){
int n=L.size();
sort(L.begin(),L.end());//将所有半平面按照极角排序。
int first,last;
vector<Point> p(n);
vector<Line> q(n);
vector<Point> ans;
q[first=last=0]=L[0];
for(int i=1;i<n;i++){
while(first<last&&!OnLeft(L[i],p[last-1]))last--;//删除顶部的半平面
while(first<last&&!OnLeft(L[i],p[first]))first++;//删除底部的半平面
q[++last]=L[i];//将当前的半平面假如双端队列顶部。
if(fabs(Cross(q[last].v,q[last-1].v))<eps){//对于极角相同的,选择性保留一个。
last--;
if(OnLeft(q[last],L[i].p))q[last]=L[i];
}
if(first<last)p[last-1]=GetLineIntersection(q[last-1],q[last]);//计算队列顶部半平面交点。
}
while(first<last&&!OnLeft(q[first],p[last-1]))last--;//删除队列顶部的无用半平面。
if(last-first<=1)return ans;//半平面退化
p[last]=GetLineIntersection(q[last],q[first]);//计算队列顶部与首部的交点。
for(int i=first;i<=last;i++)ans.push_back(p[i]);//将队列中的点复制。
return ans;
}
double PolyArea(vector<Point> p){
int n=p.size();
double ans=0;
for(int i=1;i<n-1;i++)
ans+=Cross(p[i]-p[0],p[i+1]-p[0]);
return fabs(ans)/2;
}
struct Circle
{
Point c;
double r;
Circle(){}
Circle(Point c, double r):c(c), r(r){}
Point point(double a) //根据圆心角求点坐标
{
return Point(c.x+cos(a)*r, c.y+sin(a)*r);
}
};
bool InCircle(Point x,Circle c){
return dcmp(c.r-Length(c.c-x))>=0;
}
bool OnCircle(Point x,Circle c){
return dcmp(c.r-Length(c.c-x))==0;
}
int getSegCircleIntersection(Line L,Circle C,Point *sol){
Point nor=Normal(L.v);
Line p1=Line(C.c,nor);
Point ip=GetLineIntersection(p1,L);
double dis=Length(ip-C.c);
if(dcmp(dis-C.r)>0)return 0;
Point dxy=vecunit(L.v)*sqrt(C.r*C.r-dis*dis);
int ret=0;
sol[ret]=ip+dxy;
if(OnSegment(sol[ret],L.p,L.point(1)))ret++;
sol[ret]=ip-dxy;
if(OnSegment(sol[ret],L.p,L.point(1)))ret++;
return ret;
}
double SegCircleArea(Circle C,Point a,Point b){
double a1=angle(a-C.c);
double a2=angle(b-C.c);
double da=fabs(a1-a2);
if(da>pi)da=pi*2-da;
return dcmp(Cross(b-C.c,a-C.c))*da*C.r*C.r/2.0;
}
double PolyCircleArea(Circle C,Point *p,int n){
double ret=0;
Point sol[2];
p[n]=p[0];
for(int i=0;i<n;i++){
double t1,t2;
int cnt=getSegCircleIntersection(Line(p[i],p[i+1]-p[i]),C,sol);
if(cnt==0){
if(!InCircle(p[i],C)||!InCircle(p[i+1],C))ret+=SegCircleArea(C,p[i],p[i+1]);
else ret+=Cross(p[i+1]-C.c,p[i]-C.c)/2;
}
if(cnt==1){
if(InCircle(p[i],C)&&!InCircle(p[i+1],C))ret+=Cross(sol[0]-C.c,p[i]-C.c)/2,ret+=SegCircleArea(C,sol[0],p[i+1]);
else ret+=SegCircleArea(C,p[i],sol[0]),ret+=Cross(p[i+1]-C.c,sol[0]-C.c)/2;
}
if(cnt==2){
if((p[i]<p[i+1])^(sol[0]<sol[1]))swap(sol[0],sol[1]);
ret+=SegCircleArea(C,p[i],sol[0]);
ret+=Cross(sol[1]-C.c,sol[0]-C.c)/2;
ret+=SegCircleArea(C,sol[1],p[i+1]);
}
}
return fabs(ret);
}
pair<Point,Point> pp[2200];
Point p[100010];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int T,n;
cin>>T;
double R;
Point dd;
for(int t=1;t<=T;t++){
scanf("%lf%d",&R,&n);
Point a,b;
vector<Line> L;
Line s;
a=Point(-20000,-20000);b=Point(20000,-20000);s=Line(a,b-a);L.push_back(s);
a=Point(20000,-20000);b=Point(20000,20000);s=Line(a,b-a);L.push_back(s);
a=Point(20000,20000);b=Point(-20000,20000);s=Line(a,b-a);L.push_back(s);
a=Point(-20000,10000);b=Point(-20000,-20000);s=Line(a,b-a);L.push_back(s);
for(int i=0;i<n;i++)
scanf("%lf%lf%lf%lf",&pp[i].first.x,&pp[i].first.y,&pp[i].second.x,&pp[i].second.y);
scanf("%lf%lf",&dd.x,&dd.y);
for(int i=0;i<n;i++){
a=pp[i].first;
b=pp[i].second;
if(OnLeft(Line(a,b-a),dd))L.push_back(Line(a,b-a));
else L.push_back(Line(b,a-b));
}
vector<Point> ff=HPI(L);
n=ff.size();
for(int i=0;i<n;i++)p[i]=ff[i];
Circle C(Point(0,0),R);
double ret=R*R*pi;
double ans=PolyCircleArea(C,p,n);
ans=100*ans/ret;
printf("Case %d: %.5lf%%\n",t,ans);
}
return 0;
}
HDU 3982 半平面交+圆与多边形面积交,布布扣,bubuko.com
标签:des style blog class code java
原文地址:http://blog.csdn.net/xianxingwuguan1/article/details/25425097