POJ 1474 Video Surveillance 半平面交/多边形核是否存在

http://poj.org/problem?id=1474

解法同POJ 1279 A一送一 缺点是还是O(n^2) ...nlogn的过几天补上...

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define EPS         1e-8
#define MAXN        10005
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define INF         ((1LL)<<50)
#define max(a,b)    ((a) > (b) ? (a) : (b))
#define min(a,b)    ((a) < (b) ? (a) : (b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define BUG         cout<<"BUG! "<<endl
#define LLINE        cout<<"------------------"<<endl
#define L(t)        (t << 1)
#define R(t)        (t << 1 | 1)
#define Mid(a,b)    ((a + b) >> 1)
#define lowbit(a)   (a & -a)
#define FIN         freopen("in.txt","r",stdin)
#pragma comment     (linker,"/STACK:102400000,102400000")

// typedef long long LL;
// typedef unsigned long long ULL;
// typedef __int64 LL;
// typedef unisigned __int64 ULL;
// int gcd(int a,int b){ return b?gcd(b,a%b):a; }
// int lcm(int a,int b){ return a*b/gcd(a,b); }

/*********************   F   ************************/
struct POINT{
    double x,y;
    POINT(double _x = 0, double _y = 0):x(_x),y(_y){}
}p[MAXN],q[MAXN],t[MAXN];
int n;
struct LINE{
     double a,b,c;
     POINT A,B;
     LINE(POINT _a, POINT _b):A(_a),B(_b){
        a=B.y-A.y;
        b=A.x-B.x;
        c=B.x*A.y-A.x*B.y;
     }
};
double multiply(POINT sp,POINT ep,POINT op){                //叉积 左+ 右-
    return (sp.x-op.x) * (ep.y-op.y) - (ep.x-op.x) * (sp.y-op.y);
}
POINT Intersection(LINE a,LINE b){                          //直线交点
    double u = fabs(b.A.x * a.a + b.A.y * a.b + a.c);
    double v = fabs(b.B.x * a.a + b.B.y * a.b + a.c);
    POINT t;
    t.x = (b.A.x * v + b.B.x * u) / (u + v);
    t.y = (b.A.y * v + b.B.y * u) / (u + v);
    return t;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("outm.txt","w",stdout);
    int ct = 1;
    while(cin>>n && n){
        for(int i = 0 ; i < n ; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        //暴力对每一个向量作半平面交 ...即将右侧的点和与其他直线的交点加入集合
        for(int i = 0 ; i < n ; i++) q[i] = p[i];
        int cnt = n;
        for(int i = 0 ; i < n ; i++){
            int c = 0;
            for(int j = 0 ; j < cnt ; j++){
                //点在右侧
                if(multiply(p[i],p[(i+1)%n],q[j]) <= EPS) {
                    t[c++] = q[j];
                }else {     //点在左侧,但是前后线段和该直线有交点
                    //这个顺序不要写反,否则不是顺时针会WA
                    if(multiply(p[i],p[(i+1)%n],q[(j-1+cnt)%cnt]) < -EPS){
                        t[c++] = Intersection(LINE(p[i],p[(i+1)%n]) , LINE(q[j],q[(j-1+cnt)%cnt]));
                    }
                    if(multiply(p[i],p[(i+1)%n],q[(j+1)%cnt]) < -EPS){
                        t[c++] = Intersection(LINE(p[i],p[(i+1)%n]) , LINE(q[j],q[(j+1)%cnt]));
                    }
                }
            }
            for(int j = 0 ; j < c ; j++) {q[j] = t[j];}
            cnt = c;
        }
        cout<<"Floor #"<<ct++<<endl;
        if(cnt != 0) cout<<"Surveillance is possible.
"<<endl;
        else cout<<"Surveillance is impossible.
"<<endl;
    }
    return 0;
}