poj 3335 Rotating Scoreboard（半平面交）

Posted 2020-07-08 Przz

Rotating Scoreboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6420		Accepted: 2550

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 ... x_n y_n where n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers x_i y_i 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

 

/*
poj 3335 Rotating Scoreboard（半平面交）
给一个图形,判断是否存在一个位置能够观察到图形内所有的位置
即一个图形的核
输入是顺时针,需要倒一下

hhh-2016-05-08 21:18:33
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
using namespace std;
const int  maxn = 1010;
const double PI = 3.1415926;
const double eps = 1e-8;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0)
        return -1;
    else
        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;
    }
};

struct Line
{
    Point s,t;
    double k;
    Line() {}
    Line(Point _s,Point _t)
    {
        s = _s;
        t = _t;
        k = atan2(t.y-s.y,t.x-s.x);
    }
    Point operator &(const Line &b) const
    {
        Point res = s;
        double ta = ((s-b.s)^(b.s-b.t))/((s-t)^(b.s-b.t));
        res.x += (t.x-s.x)*ta;
        res.y += (t.y-s.y)*ta;
        return res;
    }
};

bool HPIcmp(Line a,Line b)
{
    if(fabs(a.k-b.k) > eps) return a.k<b.k;
    return ((a.s-b.s)^(b.t-b.s)) < 0;
}
Line li[maxn];
void HPI(Line line[],int n,Point res[],int &resn)
{
    int tot =n;
    sort(line,line+n,HPIcmp);
    tot = 1;
    for(int i = 1; i < n; i++)
    {
        if(fabs(line[i].k - line[i-1].k) > eps)
            line[tot++] = line[i];
    }
    int head = 0,tail = 1;
    li[0] = line[0];
    li[1] = line[1];
    resn = 0;
    for(int i = 2; i < tot; i++)
    {
        if(fabs((li[tail].t-li[tail].s)^(li[tail-1].t-li[tail-1].s)) < eps||
                fabs((li[head].t-li[head].s)^(li[head+1].t-li[head+1].s)) < eps)
            return;
        while(head < tail && (((li[tail] & li[tail-1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
            tail--;
        while(head < tail && (((li[head] & li[head+1]) - line[i].s) ^ (line[i].t-line[i].s)) > eps)
            head++;
        li[++tail] = line[i];
    }
    while(head < tail && (((li[tail] & li[tail-1]) - li[head].s) ^ (li[head].t-li[head].s)) > eps)
        tail--;
    while(head < tail && (((li[head] & li[head-1]) - li[tail].s) ^ (li[tail].t-li[tail].t)) > eps)
        head++;
    if(tail <= head+1)
    return;
    for(int i = head;i < tail;i++)
        res[resn++] = li[i]&li[i+1];
    if(head < tail-1)
        res[resn++] = li[head]&li[tail];
}


Point lis[maxn];
Line line[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i < n;i++)
        {
            scanf("%lf%lf",&lis[i].x,&lis[i].y);
        }
        reverse(lis,lis+n);
        int ans;
        for(int i = 0;i < n;i++)
        {
            line[i] = Line(lis[i],lis[(i+1)%n]);
        }
        HPI(line,n,lis,ans);
        if(ans)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}