POJ1228 稳定凸包

对于共线凸包，暂时没有找到一种比较好的实现方法。

　　本题对于共线是直接O(n^2)的方法直接刷一遍的，判断点在凸包边上的个数是否<3则NO，否则YES

　　要注意到，对于一条直线，要特判。（不过按理说庄园怎么能够是一条直线呢？）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <map>
#include <deque>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define INF 0x3f3f3f3f
#define MAXN 2005

using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);
int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0) return -1;
    return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x, double _y): x(_x),y(_y) {}
    Point operator -(const Point &B) const
    {
        return Point(x-B.x, y-B.y);
    }
    Point operator +(const Point &B) const //向量相加
    {
        return Point(x+B.x, y+B.y);
    }
    double operator ^(const Point &B) const //叉积
    {
        return x*B.y - y*B.x;
    }
    double operator *(const Point &B) const //点积
    {
        return x*B.x + y*B.y;
    }
    bool operator ==(const Point &B) const
    {
        return fabs(B.x-x)<eps && fabs(B.y-y)<eps;
    }
    bool operator !=(const Point &B) const
    {
        return !((*this) == B);
    }
    double norm()//向量的模
    {
        return sqrt(x*x+y*y);
    }
    void transXY(double B) //绕原点逆时针旋转B弧度
    {
        double tx = x, ty = y;
        x = tx*cos(B) - ty*sin(B);
        y = tx*sin(B) + ty*cos(B);
    }
    void input() //读入只能用double读入
    {
        scanf("%lf%lf",&x,&y);
    }
};

struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s, Point _e)
    {
        s=_s; e=_e;
    }
};

double dist(Point a, Point b)
{
    return sqrt((a-b)*(a-b));
}

//判断点在线段上
bool OnS(Point A, Line a)
{
    return
        sgn((a.s-A)^(a.e-A)) == 0 &&
        sgn((A.x-a.s.x)*(A.x-a.e.x)) <= 0 &&
        sgn((A.y-a.s.y)*(A.y-a.e.y)) <= 0;
}

//求凸包 Graham算法
//点的编号0~n-1
//返回凸包结果Stack[0~top-1]为凸包的编号
//一个点或两个点 则凸包为一或二个点
int Stack[MAXN],top;
Point vertex[MAXN];
bool Graham_cmp(Point A, Point B)
{
    double tmp=(A-vertex[0])^(B-vertex[0]);
    if(sgn(tmp) > 0) return 1;
    if(sgn(tmp) == 0 && sgn(dist(A,vertex[0])-dist(B,vertex[0])) <= 0) return 1;
    return 0;
}
void Graham(int n)
{
    int k=0;
    for(int i=1; i<n; i++)
        if((vertex[k].y>vertex[i].y) || (vertex[k].y==vertex[i].y && vertex[k].x>vertex[i].x))
            k=i;
    swap(vertex[0], vertex[k]);
    sort(vertex+1, vertex+n, Graham_cmp);
    if(n == 1)
    {
        top=1;
        Stack[0]=0;
        return;
    }
    if(n == 2)
    {
        top=2;
        Stack[0]=0;
        Stack[1]=1;
        return;
    }
    Stack[0]=0;
    Stack[1]=1;
    top=2;
    for(int i=2; i<n; i++)
    {
        while(top > 1 && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) <= 0)
            top--;
        Stack[top++]=i;
    }
}

int main()
{
    int tt,n;
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            vertex[i].input();
        Graham(n);
        bool flag=1;
        for(int i=0; i<top; i++)
        {
            int num=0;
            for(int j=0; j<n; j++)
                if(OnS(vertex[j],Line(vertex[Stack[i]],vertex[Stack[(i+1)%top]])))
                    num++;
            if(num<3) 
            {
                flag=0;
                break;
            }
        }
        if(flag && top>=3) printf("YES
");
        else printf("NO
");
    }
    return 0;
}

加个 水平序凸包：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <utility>
#include <stack>
#include <queue>
#include <map>
#include <deque>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)<(y)?(x):(y))
#define INF 0x3f3f3f3f
#define MAXN 2005

using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);
int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0) return -1;
    return 1;
}
struct Point
{
    double x,y;
    Point(){}
    Point(double _x, double _y): x(_x),y(_y) {}
    Point operator -(const Point &B) const
    {
        return Point(x-B.x, y-B.y);
    }
    Point operator +(const Point &B) const //向量相加
    {
        return Point(x+B.x, y+B.y);
    }
    double operator ^(const Point &B) const //叉积
    {
        return x*B.y - y*B.x;
    }
    double operator *(const Point &B) const //点积
    {
        return x*B.x + y*B.y;
    }
    bool operator ==(const Point &B) const
    {
        return fabs(B.x-x)<eps && fabs(B.y-y)<eps;
    }
    bool operator !=(const Point &B) const
    {
        return !((*this) == B);
    }
    double norm()//向量的模
    {
        return sqrt(x*x+y*y);
    }
    void transXY(double B) //绕原点逆时针旋转B弧度
    {
        double tx = x, ty = y;
        x = tx*cos(B) - ty*sin(B);
        y = tx*sin(B) + ty*cos(B);
    }
    void input() //读入只能用double读入
    {
        scanf("%lf%lf",&x,&y);
    }
};

struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s, Point _e)
    {
        s=_s; e=_e;
    }
};

double dist(Point a, Point b)
{
    return sqrt((a-b)*(a-b));
}

//判断点在线段上
bool OnS(Point A, Line a)
{
    return
        sgn((a.s-A)^(a.e-A)) == 0 &&
        sgn((A.x-a.s.x)*(A.x-a.e.x)) <= 0 &&
        sgn((A.y-a.s.y)*(A.y-a.e.y)) <= 0;
}

//求凸包 Graham算法
//点的编号0~n-1
//返回凸包结果Stack[0~top-1]为凸包的编号
//一个点或两个点 则凸包为一或二个点
int Stack[MAXN],top;
Point vertex[MAXN];
bool Graham_cmp(Point A, Point B)
{
    return A.y<B.y || (A.y == B.y && A.x<B.x);
}
void Graham(int n)
{
    sort(vertex, vertex+n, Graham_cmp);
    top=0;
    for(int i=0; i<n; i++)
    {
        while(top > 1 && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) <= 0)
            top--;
        Stack[top++]=i;
    }
    int tmp=top;
    for(int i=n-2; i>=0; i--)
    {
        while(top > tmp && sgn((vertex[Stack[top-1]]-vertex[Stack[top-2]])^(vertex[i]-vertex[Stack[top-2]])) <= 0)
            top--;
        Stack[top++]=i;
    }
    if(n>1) top--;
}
int main()
{
    int tt,n;
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            vertex[i].input();
        Graham(n);

        bool flag=1;
        for(int i=0; i<top; i++)
        {
            int num=0;
            for(int j=0; j<n; j++)
                if(OnS(vertex[j],Line(vertex[Stack[i]],vertex[Stack[(i+1)%top]])))
                    num++;
            if(num<3) 
            {
                flag=0;
                break;
            }
        }
        if(flag && top>=3) printf("YES
");
        else printf("NO
");
    }
    return 0;
}