| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 4899 | Accepted: 1946 |
Description
This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the edges of the polygon. We want to place a rotating scoreboard somewhere in the hall such that a spectator sitting anywhere on the boundary of the hall can view the scoreboard (i.e., his line of sight is not blocked by a wall). Note that if the line of sight of a spectator is tangent to the polygon boundary (either in a vertex or in an edge), he can still view the scoreboard. You may view spectator‘s seats as points along the boundary of the simple polygon, and consider the scoreboard as a point as well. Your program is given the corners of the hall (the vertices of the polygon), and must check if there is a location for the scoreboard (a point inside the polygon) such that the scoreboard can be viewed from any point on the edges of the polygon.
Input
The first number in the input line, T is the number of test cases. Each test case is specified on a single line of input in the form n x1 y1 x2 y2 ... xn yn where n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers xi yi sequence specify the vertices of the polygon sorted in order.
Output
The output contains T lines, each corresponding to an input test case in that order. The output line contains either YES or NO depending on whether the scoreboard can be placed inside the hall conforming to the problem conditions.
Sample Input
2 4 0 0 0 1 1 1 1 0 8 0 0 0 2 1 2 1 1 2 1 2 2 3 2 3 0
Sample Output
YES NO
直接半平面交，假如半平面为空输出NO，否则输出YES，很水。
代码：
/* ***********************************************
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 10000000
#define eps 1e-8
#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));
}
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;
}
};
bool OnLeft(const Line &L,const Point &p){
return dcmp(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。
/*for(int i=0;i<n;i++){
cout<<"han "<<i<<" ";
printf("%.2lf %.2lf %.2lf %.2lf %.2lf\n",L[i].p.x,L[i].p.y,L[i].v.x,L[i].v.y,L[i].ang);
}*/
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;
}
Point pp[200],v[200],v2[200];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n,T;
cin>>T;
while(T--){
cin>>n;
for(int i=0;i<n;i++)scanf("%lf%lf",&pp[i].x,&pp[i].y);
reverse(pp,pp+n);
vector<Line> L;
for(int i=0;i<n;i++)
L.push_back(Line(pp[i],pp[(i+1)%n]-pp[i]));
vector<Point> ans=HPI(L);
if(ans.size())puts("YES");
else puts("NO");
}
return 0;
}
POJ 3335 半平面交求多边形的核,布布扣,bubuko.com
原文地址：http://blog.csdn.net/xianxingwuguan1/article/details/25462637
