poj3335 Rotating Scoreboard半平面交

Posted 2022-11-23

题目链接：​​http://poj.org/problem?id=3335​​​
题意：就是有个球场，球场的形状是个凸多边形，然后观众是坐在多边形的边上的，问你是否在球场上有个地方可以放一个记分牌，然后所有的观众都可以看得到的
解析：也就是相当于求凸多边形的核，半平面交，裸的模板题

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
const int maxn = 1e5+100;
const double eps = 0;
const double pi = acos(-1.0);
int dblcmp(double x)

    if(fabs(x)<eps)
        return 0;
    return x>0?1:-1;

struct point

    double x,y;
    point() ;
    point(double _x,double _y)
    
        x = _x;
        y = _y;
    
;
//半平面相交（直线切割多边形）（点从标号1开始）
point points[maxn],p[maxn],q[maxn];
int cCnt,curCnt;
void getline(point s,point e,double &a,double &b,double &c)

    //获取直线ax+by+c=0
    a = e.y-s.y;
    b = s.x-e.x;
    c = e.x*s.y-e.y*s.x;

void init(int n)

    for(int i=1;i<=n;i++)
        p[i] = points[i];
    p[n+1] = p[1];
    p[0] = p[n];
    cCnt = n;

point cross(point p1,point p2,double a,double b,double c)

    //求直线p1p2与已知直线ax+by+c=0的交点
    double u = fabs(a*p1.x+b*p1.y+c);
    double v = fabs(a*p2.x+b*p2.y+c);
    return point((p1.x*v+p2.x*u)/(u+v),(p1.y*v+p2.y*u)/(u+v));

double x_mul(point p0,point p1,point p2)

    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);

void cut(double a,double b,double c)

    curCnt = 0;
    for(int i=1;i<=cCnt;i++)
    
        if(a*p[i].x+b*p[i].y+c>=eps)
            q[++curCnt] = p[i];
        else
        
            if(a*p[i-1].x+b*p[i-1].y+c > eps)
                q[++curCnt] = cross(p[i],p[i-1],a,b,c);
            if(a*p[i+1].x+b*p[i+1].y+c > eps)
                q[++curCnt] = cross(p[i],p[i+1],a,b,c);
        
    
    for(int i=1;i<=curCnt;i++)
        p[i] = q[i];
    p[curCnt+1] = q[1];
    p[0] = p[curCnt];
    cCnt = curCnt;

void slove(int n)

    //注意：默认点是顺时针，如果题目不是顺时针，规整化方向
    init(n);
    for(int i=1;i<=n;i++)
    
        double a,b,c;
        getline(points[i],points[i+1],a,b,c);
        cut(a,b,c);
    

int main(void)

    int t;
    cin>>t;
    while(t--)
    
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf %lf",&points[i].x,&points[i].y);
        points[n+1] = points[1];
        slove(n);
        if(cCnt>=1)
            puts("YES");
        else
            puts("NO");
    
    return 0;