POJ

pro：顺时针给定多边形，问是否可以放一个监控，可以监控到所有地方，即问是否存在多边形的核。 此题如果两点在同一边界上（且没有被隔段），也可以相互看到。

sol：求多边形是否有核。先给直线按角度排序，然后增量法即可，复杂度O(NlogN)。

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstdio>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=10010;
const double eps=1e-12;
struct point{
    double x,y;
    point(){}
    point(double xx,double yy):x(xx),y(yy){}
};
struct line{
    point a;//起点
    point p;//起点到终点的向量
    double angle;
};
double dot(point a,point b){ return a.x*b.x+a.y*b.y;}
double det(point a,point b){ return a.x*b.y-a.y*b.x;}
point operator *(point A,double p){ return point(A.x*p,A.y*p);}
point operator +(point A,point B){return point(A.x+B.x,A.y+B.y);}
point operator -(point A,point B){return point(A.x-B.x,A.y-B.y);}
double getangle(point a){ return atan2(a.y,a.x);}
double getangle(line a){ return getangle(a.p);}
point llintersect(line A,line B)
{
    point C=A.a-B.a;
    double t=det(C,B.p)/det(B.p,A.p);
    return A.a+A.p*t;
}
point s[maxn]; line t[maxn],q[maxn]; int head,tail;
bool cmp(line a,line b){
    double A=getangle(a),B=getangle(b);
    point t=(b.a+b.p)-a.a;
    if(fabs(A-B)<eps) return det(a.p,t)>=0.0;
    return A<B;
}
bool onright(line P,line a,line b)
{
    point o=llintersect(a,b);
    point Q=o-P.a;
    return det(Q,P.p)>0; //如果同一直线上不能相互看到，则>=0
}
bool halfplaneintersect(int N)
{
    s[N+1]=s[1];
    rep(i,1,N) t[i].a=s[i],t[i].p=s[i+1]-s[i];
    sort(t+1,t+N+1,cmp);
    int tot=0;
    rep(i,1,N-1) {
        if(fabs(getangle(t[i])-getangle(t[i+1]))>eps)
          t[++tot]=t[i];
    }
    t[++tot]=t[N]; head=tail=0;
    rep(i,1,tot){
        while(tail>head+1&&onright(t[i],q[tail],q[tail-1])) tail--;
        while(tail>head+1&&onright(t[i],q[head+1],q[head+2])) head++;
        q[++tail]=t[i];
    }
    while(tail>head+1&&onright(t[head+1],q[tail],q[tail-1])) tail--;return tail-head>2;
}
void solve(int C,int N)
{
    for(int i=N;i>=1;i--) scanf("%lf%lf",&s[i].x,&s[i].y);
    printf("Floor #%d
",C);
    if(halfplaneintersect(N)) puts("Surveillance is possible.
");
    else puts("Surveillance is impossible.
");
}
int main()
{
    int N,C=0;
    while(~scanf("%d",&N)&&N) solve(++C,N);
    return 0;
}