Rotating Scoreboard——poj3335

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 nx1y1x2y2 ... xnynwhere n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers xiyi 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

 

求两点的直线方程
y=kx+m,
y1=kx1+m
y2=kx2+m,
k=(y2-y1)/(x2-x1)
m=y1-(y2-y1)*x1/(x2-x1)
y=(y2-y1)/(x2-x1)x+(y1*(x2-x1)-x1(y2-y1))/(x2-x1)
 (y1-y2)x+(x2-x1)y+(x1y2-x2y1)=0,
a=y1-y2,
b=x2-x1,
c=x1y2-x2y1,

#include <iostream>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
struct haha{
    double x,y;
};
int n,m;//n的原先的点数,m是新切割出的多边形的点数
haha we[105];//记录最开始的多边形
haha q[105]; //临时保存新切割的多边形
haha p[105]; //保存新切割出的多边形
double a,b,c;
//double e,f,g; 
void get(haha x,haha y){//得到直线ax+by+c=0
    a=y.y-x.y;
    b=x.x-y.x;
    c=y.x*x.y-x.x*y.y;
}

haha intersect(haha x,haha y){//得到直线ax+by+c==0与点x和y所连直线的交点
    double u=fabs(a*x.x+b*x.y+c);
    double v=fabs(a*y.x+b*y.y+c);
    haha ans;
    ans.x=(x.x*v+y.x*u)/(u+v);
    ans.y=(x.y*v+y.y*u)/(u+v);
    return ans;
}

void cut(){//切割多边形
    int d=0;
    for(int i=1;i<=m;i++){//题目是顺时钟给出的点,点在直线右边的话，那么带入值就会大于等于0
        if(a*p[i].x+b*p[i].y+c>=0){ //说明这个点还在切割后的多边形内，将其保留
            q[++d]=p[i];            
        }
        else{
            if(a*p[i-1].x+b*p[i-1].y+c>0){//该点不在多边形内，但是它和它相邻的点构成直线也需判断			 
                q[++d]=intersect(p[i-1],p[i]); //所以保留交点
            }
            if(a*p[i+1].x+b*p[i+1].y+c>0){
                q[++d]=intersect(p[i+1],p[i]);
            }
        }
    }
    m=d;
    for(int i=1;i<=d;i++){
        p[i]=q[i];
    }
    p[d+1]=q[1];
    p[0]=q[d];
}

void solve(){
    for(int i=1;i<=n;i++){
        p[i]=we[i];
    }
    we[n+1]=we[1];
    p[n+1]=p[1];
    p[0]=p[n];
    m=n;
    for(int i=1;i<=n;i++){
        get(we[i],we[i+1]); //使用原来的点确定直线ax+by+c==0
        cut();  				//用直线ax+by+c==0切割多边形
    }
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lf%lf",&we[i].x,&we[i].y);
        }
        solve();
        if(m==0){
            printf("NO\n");
        }
        else{
            printf("YES\n");
        }
    }
    return 0;
}