Video Surveillance POJ

Video Surveillance

 POJ - 1474 

题意：多边形是否有内核

思路：平面半交题，注意直线的存入顺序

// 
// Created by HJYL on 2020/2/6.
//
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double db;
const int N=207;
const db eps=1e-7;
int sign(db k){if(k>eps) return 1;if(k<-eps) return -1; return 0;}
int dcmp(db k1,db k2){return sign(k1-k2);}
struct point{
    db x,y;
    point operator - (const point &k)const{
        return (point){x-k.x,y-k.y};
    }
    point operator + (const point &k)const{
        return (point){x+k.x,y+k.y};
    }
    point operator * (const db &k)const{
        return (point){x*k,y*k};
    }
    point operator / (const db &k)const{
        return (point){x/k,y/k};
    }
    db angle(){return atan2(y,x);}
    void print(){printf("(%f,%f)
",x,y);}
}P[N];
db cross(point k1,point k2){
    return k1.x*k2.y-k1.y*k2.x;
}
db dot(point k1,point k2){
    return k1.x*k2.x+k1.y*k2.y;
}
struct line{
    point s,e;
    db angle(){return (e-s).angle();}
}L[N],dq[N];
point getLL(line k1,line k2){
    db w1=cross(k1.s-k2.s,k2.e-k2.s),w2=cross(k2.e-k2.s,k1.e-k2.s);
    return (k1.s*w2+k1.e*w1)/(w1+w2);
}
bool onRight(line k,point p){
    return sign((cross(p-k.s,k.e-k.s)))>0;
}
bool cmp(line k1,line k2){
    if(dcmp(k1.angle(),k2.angle())==0) return onRight(k2,k1.s);
    return k1.angle()<k2.angle();
}
int tot;
bool halfplaneinsert(){
    sort(L+1,L+1+tot,cmp);
    int cnt=0;
    for(int i=1;i<=tot;i++){
        if(i<tot&&dcmp(L[i].angle(),L[i+1].angle())==0) continue;
        L[++cnt]=L[i];
    }
    int tail=-1,head=0;
    for(int i=1;i<=cnt;i++){
        while(tail-head>=1&&onRight(L[i],getLL(dq[tail],dq[tail-1]))) tail--;
        while(tail-head>=1&&onRight(L[i],getLL(dq[head],dq[head+1]))) head++;
        dq[++tail]=L[i];
    }
    while(tail-head>=1&&onRight(dq[head],getLL(dq[tail],dq[tail-1]))) tail--;
    while(tail-head>=1&&onRight(dq[tail],getLL(dq[head],dq[head+1]))) head++;
    return tail-head>=2;
}
int main()
{
    int cases=0;
    int n;
    while(~scanf("%d",&n)&&n) {
            for (int i = 1; i <= n; i++) {
                scanf("%lf%lf", &P[i].x, &P[i].y);
            }
            tot = 0;
            for (int i = n; i > 1; i--) {
                L[++tot] = (line) {P[i], P[i - 1]};
            }
            L[++tot] = (line) {P[1], P[n]};
            printf("Floor #%d
",++cases);
            if (halfplaneinsert()) printf("Surveillance is possible.
");
            else printf("Surveillance is impossible.
");
            printf("
");
        }
}