POJ 1228 凸包惟一性判断（多边形模板）
E - Grandpa's Estate
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Description
Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one was a piece of convex polygon shaped farm in the grandpa's birth village. The farm was originally separated from the neighboring farms by a thick rope hooked to some spikes (big nails) placed on the boundary of the polygon. But, when Kamran went to visit his farm, he noticed that the rope and some spikes are missing. Your task is to write a program to help Kamran decide whether the boundary of his farm can be exactly determined only by the remaining spikes.
Input
The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains an integer n (1 <= n <= 1000) which is the number of remaining spikes. Next, there are n lines, one line per spike, each containing a pair of integers which are x and y coordinates of the spike.
Output
There should be one output line per test case containing YES or NO depending on whether the boundary of the farm can be uniquely determined from the input.
Sample Input
1 6 0 0 1 2 3 4 2 0 2 4 5 0
Sample Output
NO
题意：判断凸包是否惟一。
思路：把凸包的结点先变成顶点，然后再判断两个顶点之间是否还有结点，如果有顶点，说明两个顶点之间的边是惟一的。如果顶点之间没有结点，所以这两个顶点之间可以有其他加入的结点让这个凸包不惟一。
#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#define MAX 111116
#define eps 1e-7
using namespace std;
int sgn(const double &x){ return x < -eps? -1 : (x > eps);}
inline double sqr(const double &x){ return x * x;}
inline int gcd(int a, int b){ return !b? a: gcd(b, a % b);}
struct Point
{
double x, y;
Point(const double &x = 0, const double &y = 0):x(x), y(y){}
Point operator -(const Point &a)const{ return Point(x - a.x, y - a.y);}
Point operator +(const Point &a)const{ return Point(x + a.x, y + a.y);}
Point operator *(const double &a)const{ return Point(x * a, y * a);}
Point operator /(const double &a)const{ return Point(x / a, y / a);}
bool operator < (const Point &a)const{ return sgn(x - a.x) < 0 || (sgn(x - a.x) == 0 && sgn(y - a.y) < 0);}
bool operator == (const Point &a)const{ return sgn(sgn(x - a.x) == 0 && sgn(y - a.y) == 0);}
friend double det(const Point &a, const Point &b){ return a.x * b.y - a.y * b.x;}
friend double dot(const Point &a, const Point &b){ return a.x * b.x + a.y * b.y;}
friend double dist(const Point &a, const Point &b){ return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}
void in(){ scanf("%lf %lf", &x, &y); }
void out()const{ printf("%lf %lf\n", x, y); }
};
struct Line //线段类
{
Point s, t;
Line() {}
Line(const Point &s, const Point &t):s(s), t(t) {}
void in() { s.in(),t.in(); }
double pointDistLine(const Point &p)
{
if(sgn(dot(t - s, p - s)) < 0)return dist(p, s);
if(sgn(dot( s - t, p - t)) < 0)return dist(p, t);
return fabs(det(t - s, p - s)) / dist(s, t);
}
bool pointOnLine(const Point &p)
{
return sgn(det(s - p, t - p)) == 0 && sgn(dot(s - p, t - p)) < 0;
}
};
struct Poly //多边形类
{
vector<Point>a;
vector<Point>p; //顺时针凸包
vector<Point>tb;// 逆时针凸包
void in(const int &r)
{
p.resize(r); //不早凸包的时候可以把p改为a
for(int i = 0; i < r; i++) p[i].in();
}
//计算多边形的周长
double perimeter()
{
double sum=0;
int n=a.size();
for(int i=0;i<n;i++) sum+=dist(a[i],a[(i+1)%n]);
return sum;
}
//计算多边形的面积
double getArea()
{
int n = tb.size(); //平常的多边形就把tb换成a
double ans=0;
for(int i = 0; i < n; i++) ans += det(tb[i], tb[(i + 1)%n]);
return ans / 2;
}
//计算多边形的重心坐标
Point getMassCenter()
{
Point center(0, 0);
if(sgn(getArea())==0) return center; //面积为0情况，当然这题说了面积不可能为0可不写
int n = a.size();
for(int i = 0; i < n; i++)
center =center + (a[i] + a[(i + 1) % n]) * det(a[i], a[(i + 1) % n]);
return center / getArea() / 6;
}
//计算点t是否在多边形内,返回0指在外，1指在内，2指在边界上
int pointin(Point t)
{
int num=0,i,d1,d2,k,n=a.size();
for(i=0;i<n;i++)
{
Line line=Line(a[i],a[(i+1)%n]);
if(line.pointOnLine(t)) return 2;
k=sgn(det(a[(i+1)%n]-a[i],t-a[i]));
d1=sgn(a[i].y-t.y);
d2=sgn(a[(i+1)%n].y-t.y);
if(k>0&&d1<=0&&d2>0) num++;
if(k<0&&d2<=0&&d1>0) num--;
}
return num!=0;
}
//计算多边形边界的格点数
int border()
{
int num=0,i,n=a.size();
for(i=0;i<n;i++)
num+=gcd(abs(int(a[(i+1)%n].x-a[i].x)),abs(int(a[(i+1)%n].y-a[i].y)));
return num;
}
//计算多边形内的格点数
//Pick公式：面积=内部格点数+边界格点数/2-1
int inside()
{
return int(getArea())+1-border()/2;
}
//判断点集是否为凸包(返回m-1==n),或者用凸包点算出凸包顶点tb(本题即是)
void isCanHull()
{
sort(p.begin(), p.end());
p.erase(unique(p.begin(), p.end()), p.end());
int n = p.size();
tb.resize(n * 2 + 5);
int m = 0;
for(int i = 0; i < n; i++)
{
while(m > 1 && sgn(det(tb[m - 1] - tb[m - 2], p[i] - tb[m - 2])) <= 0)m--;
tb[m++] = p[i];
}
int k = m;
for(int i = n - 2; i >= 0; i--)
{
while(m > k && sgn(det(tb[m - 1] - tb[m -2], p[i] - tb[m - 2])) <= 0)m--;
tb[m++] = p[i];
}
tb.resize(m);
if(m > 1)tb.resize(m - 1);
//for(int i = 0; i < m - 1; i++) tb[i].out();
}
//判断点t(圆心)是否在凸包内部，这个是O(logn)的算法
bool isContainOlogn(const Point &t)
{
int n = tb.size();
if(n < 3)return 0;
Point g = (tb[0] + tb[n / 3] + tb[n * 2 / 3] )/ 3.0;
int l = 0, r = n;
while(l + 1 < r)
{
int mid = (l + r) >> 1;
int k = sgn(det(tb[l] - g, tb[mid] - g) );
int dl = sgn(det(tb[l] - g, t - g) );
int dr = sgn(det(tb[mid] - g, t - g) );
if(k > 0)
{
if(dl >= 0 && dr < 0) r = mid;
else l = mid;
}
else
{
if(dl < 0 && dr >= 0) l = mid;
else r = mid;
}
}
r %= n;
int res = sgn(det(tb[l] - t, tb[r] - t));
if(res >= 0) return true;
return false;
}
//判断凸包是否惟一,若顶点之间有点说明惟一
bool isUnique()
{
if(sgn(getArea()) == 0) return false;
int n = tb.size();
int np = p.size();
for(int i = 0; i < n; i++)
{
Line line(tb[i], tb[(i + 1) % n]);
bool isFind = false;
for(int j = 0; j < np; j++)
{
if(line.pointOnLine(p[j]))
{
isFind = true;
break;
}
}
if(!isFind) return false;
}
return true;
}
}poly;
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
poly.in(n);
poly.isCanHull();
if(poly.isUnique()) puts("YES");
else puts("NO");
}
return 0;
}