半平面交模板 HDU 1469

题意就是要判断一个多边形是否存在核。

我们可以把沿着顺时针方向走这个多边形，对于每个边向量，我们取其右边的半平面，判断交是否为空即可。

对于半平面交算法，我只理解了O(n^2)的算法，大概就是用向量去切割多边形，对于O(nlogn)的算法，我从网上各种搜集以及参考了蓝书的实现，给出了一份能看的代码。

O(n^2)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 1005
const double eps=0.00000001;
struct Point {double x,y;};
struct Vector {double x,y;};
int Case,n,tot1,tot2;
Point a[maxn],b[maxn],tmp[maxn];

Vector operator - (Point x,Point y) 
{
    return (Vector){x.x-y.x,x.y-y.y};
}

bool operator == (Point x,Point y)
{
    return fabs(x.x-y.x)<eps&&fabs(x.y-y.y)<eps;
}

double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}

double Dot(Vector a,Vector b)
{
    return a.x*b.x+a.y*b.y;
}

void solve(void) 
{
    a[0]=a[n];tot1=0;
    b[++tot1]=(Point){-1e9,-1e9};
    b[++tot1]=(Point){-1e9,1e9};
    b[++tot1]=(Point){1e9,1e9};
    b[++tot1]=(Point){1e9,-1e9};
    for (int i=1;i<=n;i++) 
    {
        Point A=a[i];
        Point B=a[(i+1)%n];
        tot2=0;
        for (int j=1;j<=tot1;j++) 
        {
            Point C=b[j];
            Point D=(j+1)%tot1?b[(j+1)%tot1]:b[tot1];
            if (Cross(B-A,C-A)<=0) tmp[++tot2]=C;
            if (fabs(Cross(B-A,C-D))>eps) 
            {
                Vector v=A-C;
                double lim=Cross(D-C,v)/Cross(B-A,D-C);
                tot2++;
                tmp[tot2].x=A.x+lim*(B-A).x;
                tmp[tot2].y=A.y+lim*(B-A).y;
                if (tmp[tot2]==C||tmp[tot2]==D) tot2--;
                else if (Dot(tmp[tot2]-C,tmp[tot2]-D)>0) tot2--; 
            }
        }
        tot1=tot2;
        for (int i=1;i<=tot1;i++) b[i]=tmp[i];
    }
    Case++;
    printf("Floor #%d\n",Case);
    tot1?puts("Surveillance is possible."):puts("Surveillance is impossible.");
    puts("");
    //for (int i=1;i<=tot1;i++) printf("%.3f %.3f\n",b[i].x,b[i].y);
}

int main()
{
    while (1) 
    {
        scanf("%d",&n);
        if (n==0) break;
        for (int i=1;i<=n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
        solve();
    }
    return 0;
}

O(nlogn)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 100005
const double eps=0.00000001;
struct Point {double x,y;};
struct Vector {double x,y;};
struct Line{Point p;Vector v;double ang;};
Point a[maxn],b[maxn];
Line c[maxn],dque[maxn];
int n,Case;

Vector operator - (Point x,Point y)
{
    return (Vector){x.x-y.x,x.y-y.y};
}

double Cross(Vector a,Vector b)
{
    return a.x*b.y-a.y*b.x;
}

inline bool cmp(Line x,Line y)
{
    if (fabs(x.ang-y.ang)>eps) return x.ang<y.ang;
    return Cross(x.v,y.p-x.p)>0;
}

bool check(Line A,Line B,Line C) 
{
    Vector v=A.p-B.p;
    double lim=Cross(B.v,v)/Cross(A.v,B.v);
    Point tmp;
    tmp.x=A.p.x+lim*A.v.x;
    tmp.y=A.p.y+lim*A.v.y;
    return Cross(C.v,tmp-C.p)>0;
}

int main()
{
    while (1) 
    {
        scanf("%d",&n);
        if (n==0) break;
        for (int i=1;i<=n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
        a[0]=a[n];
        for (int i=1;i<=n;i++) 
        {
            c[i].p=a[i];
            c[i].v=a[(i+1)%n]-a[i];
            c[i].ang=atan2(c[i].v.y,c[i].v.x);
            //printf("%.3f\n",c[i].ang);
        }
        sort(c+1,c+n+1,cmp);
        int nn=1;
        for (int i=2;i<=n;i++)  
            if (fabs(c[nn].ang-c[i].ang)>eps) c[++nn]=c[i];
        n=nn;
        //printf("%d\n",n);
        dque[1]=c[1];
        dque[2]=c[2];
        int l=1,r=2;
        for (int i=3;i<=n;i++) 
        {
            while (l<r&&check(dque[r],dque[r-1],c[i])) r--;
            while (l<r&&check(dque[l],dque[l+1],c[i])) l++;
            dque[++r]=c[i];
        }
        while (l<r&&check(dque[r],dque[r-1],dque[l])) r--;
        while (l<r&&check(dque[l],dque[l+1],dque[r])) l++;
        printf("Floor #%d\n",++Case);
        r-l>1?puts("Surveillance is possible."):puts("Surveillance is impossible.");
        puts("");
    }
    return 0;
}