題意看了很久才懂。。
判斷給定的凸包中的某些點能否唯一確定一個原凸多邊形,可以推知如果凸包上相鄰兩頂點之間沒有其他頂點的話,則不能唯一確定凸多邊形(類似於兩點確定一條直線)
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=1100;
const double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x = _x;y = _y;
}
//向量
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
//叉積
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
//點積
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
bool operator ==(const Point &b)const
{
return !sgn(x - b.x) && !sgn(y - b.y);
}
bool operator !=(const Point &b)const
{
return sgn(x - b.x) || sgn(y - b.y);
}
void input()
{
scanf("%lf%lf",&x,&y);
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//兩直線相交求交點
//第一個值爲0表示直線重合,爲1表示平行,爲0表示相交,爲2是相交
//只有第一個值爲2時,交點纔有意義
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == 0)
{
if(sgn((s-b.e)^(b.s-b.e)) == 0)
return make_pair(0,res);//重合
else return make_pair(1,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(2,res);
}
};
bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P)) == 0 &&
sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&
sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
}
Point list[MAXN];
int stack[MAXN],top;
int cross(Point p0,Point p1,Point p2) //計算叉積 p0p1 X p0p2
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
double dis(Point p1,Point p2) //計算 p1p2的 距離
{
return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
bool cmp(Point p1,Point p2) //極角排序函數 , 角度相同則距離小的在前面
{
int tmp=cross(list[0],p1,p2);
if(tmp>0) return true;
else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;
else return false;
}
void init(int n) //輸入,並把 最左下方的點放在 list[0] 。並且進行極角排序
{
int i,k;
Point p0;
scanf("%lf%lf",&list[0].x,&list[0].y);
p0.x=list[0].x;
p0.y=list[0].y;
k=0;
for(i=1;i<n;i++)
{
scanf("%lf%lf",&list[i].x,&list[i].y);
if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].x)) )
{
p0.x=list[i].x;
p0.y=list[i].y;
k=i;
}
}
list[k]=list[0];
list[0]=p0;
sort(list+1,list+n,cmp);
}
void graham(int n)
{
int i;
if(n==1) {top=0;stack[0]=0;}
if(n==2)
{
top=1;
stack[0]=0;
stack[1]=1;
}
if(n>2)
{
for(i=0;i<=1;i++) stack[i]=i;
top=1;
for(i=2;i<n;i++)
{
while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) top--;
top++;
stack[top]=i;
}
}
}
int main()
{
int t;
scanf("%d", &t);
while (t--) {
int n;
scanf("%d", &n);
init(n);
graham(n);
if (n < 3) {
puts("NO");
continue;
}
// for (int i = 0; i <= top; i++)
// printf("%d %lf %lf\n", stack[i], list[stack[i]].x, list[stack[i]].y);
bool ok = true;
Line s0(list[0], list[1]);
bool coline = true;
for (int i = 2; i < n; i++) {
Line s1(list[0], list[i]);
if ((s0 & s1).first != 0) {
coline = false;
break;
}
}
if (coline) {
puts("NO");
continue;
}
for (int i = 0; i <= top; i++) {
Line seg(list[stack[i]], list[stack[(i+1) % (top+1)]]);
bool flag = false;
for (int j = 0; j < n; j++) {
if(list[j] != seg.s && list[j] != seg.e && OnSeg(list[j], seg)) {
flag = true;
break;
}
}
if (!flag) {
ok = false;
break;
}
}
puts(ok ? "YES" : "NO");
}
return 0;
}