POJ 1228 Grandpa's Estate 凸包 唯一性

LINK

题意：给出一个点集，问能否够构成一个稳定凸包，即加入新点后仍然不变。

思路：对凸包的唯一性判断，对任意边判断是否存在三点及三点以上共线，如果有边不满足条件则NO，注意使用水平序，这样一来共线点的包括也较为容易，而极角序对始边和终边的共线问题较为麻烦。

/** @Date    : 2017-07-17 21:08:41
  * @FileName: POJ 1228 稳定凸包.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;


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);
	}
	double operator *(const point &b) const 
	{
		return x * b.x + y * b.y;
	}
	double operator ^(const point &b) const
	{
		return x * b.y - y * b.x;
	}
};

double xmult(point p1, point p2, point p0)  
{  
    return (p1 - p0) ^ (p2 - p0);  
}  

double distc(point a, point b)
{
	return sqrt((double)((b - a) * (b - a)));
}
int sign(double x)
{
	if(fabs(x) < eps)
		return 0;
	if(x < 0)
		return -1;
	else 
		return 1;
}

////////
int n;
point stk[N];
point p[N];
int cmp(point a, point b)//以p[0]基准 极角序排序
{
	int t = xmult(a, b, p[0]);
	if(t > 0)
		return 1;
	if(t == 0)
		return distc(a, p[0]) < distc(b, p[0]);
	if(t < 0)
		return 0;
}
int cmpC(point a, point b)//水平序排序
{
	return sign(a.x - b.x) < 0 || (sign(a.x - b.x) == 0 && sign(a.y - b.y) < 0);
}
int GrahamA()
{
	double mix, miy;
	mix = miy = 1e10;
	int pos = 0;
	for(int i = 0; i < n; i++)
	{
		if(p[i].y < miy || (p[i].y == miy && p[i].x < mix))
		{
			mix = p[i].x, miy = p[i].y;
			pos = i;
		}
	}
	swap(p[0], p[pos]);
	sort(p + 1, p + n, cmp);
	int top = 0;
	stk[0] = p[0];
	stk[1] = p[1];
	for(int i = 0; i < n; i++)
	{
		while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], p[i])) < 0)
			top--;
		stk[top++] = p[i];
	}
	//stk[++top] = p[0];
	return top;
}

int Graham()//水平序
{
	sort(p, p + n, cmpC);
	int top = 0;
	for(int i = 0; i < n; i++)
	{
		while(top >= 2 && sign(xmult(stk[top - 2], stk[top - 1], p[i])) < 0)
			top--;
		stk[top++] = p[i];
	}
	int tmp = top;
	for(int i = n - 2; i >= 0; i--)
	{
		while(top > tmp && sign(xmult(stk[top - 2],stk[top - 1] ,p[i] )) < 0)
			top--;
		stk[top++] = p[i];
	}
	if(n > 1)
		top--;
	return top;
}


int check(int m)
{
	//cout << m << endl;
	for(int i = 1; i < m; i++)
	{
		//cout << i << endl;
		//cout << "x:" << stk[i].x << "y:" << stk[i].y << endl;
		//cout << xmult(stk[i - 1], stk[(i + 1)%(m)], stk[i]) << "~" << xmult(stk[i], stk[(i + 2)%(m)], stk[(i + 1)%(m)]) << endl;
		
		if(sign(xmult(stk[i - 1], stk[(i + 1)%(m)], stk[i])) != 0 
			&& sign(xmult(stk[i], stk[(i + 2)%(m)], stk[(i + 1)%(m)])) != 0)
			return 0;
	}
	return 1;
}
/////////
int main()
{
	int T;
	cin >> T;
	while(T--)
	{
		scanf("%d", &n);
		for(int i = 0; i < n; i++)
		{
			double x, y;
			scanf("%lf%lf", &x, &y);
			p[i] = point(x, y);
		}
		if(n < 6)
		{
			printf("NO
");
			continue;
		}
		int cnt = Graham();
		// for(int i = 0 ; i < cnt; i++)
		// 	cout << stk[i].x << "%" << stk[i].y << endl;
		printf("%s
", check(cnt)?"YES":"NO");

	}
    return 0;
}