poj 3335 Rotating Scoreboard（半平面交，求解多边形内核）

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本文介绍了一种算法，用于确定旋转计分板在多边形大厅内的最佳位置，确保所有边界上的观众都能看到计分板。通过多次切割多边形来求解其核，最终判断是否存在合适的放置位置。

Rotating Scoreboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7397		Accepted: 2963

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

题目大意：在场地放一个旋转的计分板，要求每个人都能看到，即每个人与计分板的连线之间都没有障碍物，所以求场地多边形的核。

代码如下：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define exp 1e-10

struct Node {
    double x;
    double y;
};

Node point[105];//记录最开始的多边形
Node q[105]; //临时保存新切割的多边形
Node p[105]; //保存新切割出的多边形
int n,m;//n的原先的点数,m是新切割出的多边形的点数
double a,b,c;

void getline(Node x,Node y)  //传入两点坐标,获取直线ax+by+c==0
{
    a=y.y-x.y;
    b=x.x-y.x;
    c=y.x*x.y-x.x*y.y;
}

Node intersect(Node x,Node y) //获取直线ax+by+c==0  和点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);
    Node 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()  //用直线ax+by+c==0切割多边形
{
    int cutm=0,i;
    for(i=1;i<=m;i++)
    {
        if(a*p[i].x+b*p[i].y+c>=0)  //题目是顺时钟给出点的
        {                           //所以一个点在直线右边的话，那么带入值就会大于等于0
            q[++cutm]=p[i];         //说明这个点还在切割后的多边形内，将其保留
        }
        else
        {
            if(a*p[i-1].x+b*p[i-1].y+c>0) //该点不在多边形内，但是如果和它相邻的点在直线右侧，则构成直线与
            {                             //ax+by+c==0所构成的交点可能在新切割出的多边形内；
                q[++cutm]=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[++cutm]=intersect(p[i+1],p[i]);
            }
        }
    }
    for(i=1;i<=cutm;i++)//转存一下，下一次割边的时候用
    {
        p[i]=q[i];
    }
    p[cutm+1]=q[1];//这两个操作不要忘记
    p[0]=q[cutm];
    m=cutm;
}

void solve()
{
    int i;
    for(i=1;i<=n;i++)
    {
        p[i]=point[i];
    }
    point[n+1]=point[1];
    p[n+1]=p[1];
    p[0]=p[n];
    m=n;
    for(i=1;i<=n;i++)
    {
        getline(point[i],point[i+1]); //根据point[i]和point[i+1]确定直线ax+by+c==0
        cut();  //用直线ax+by+c==0切割多边形
    }
}

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