首页 > 代码库 > POJ 1228 Grandpa's Estate [稳定凸包]
POJ 1228 Grandpa's Estate [稳定凸包]
Grandpa‘s Estate
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 13112 | Accepted: 3697 |
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
16 0 01 23 42 02 4 5 0
Sample Output
NO
Source
Tehran 2002 Preliminary
题意：输入一个凸包上的点（没有凸包内部的点，要么是凸包顶点，要么是凸包边上的点），判断这个凸包是否稳定。所谓稳
定就是判断能不能在原有凸包上加点，得到一个更大的凸包，并且这个凸包包含原有凸包上的所有点。
就是求凸包时把共线的点也加入凸包然后判断相邻两个连续三个点是不是至少有一个是共线的如果稳定那么必定有一个是共线还有一个特判至少六个点才行哦明白了吧我现在心情不好因为刚刚钥匙串出问题了网络出问题了所以凑合着看吧
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>using namespace std;typedef long long ll;const int N=1005;const double eps=1e-8;const double pi=acos(-1);inline int read(){ char c=getchar();int x=0,f=1; while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();} return x*f;}inline int sgn(double x){ if(abs(x)<eps) return 0; else return x<0?-1:1;}struct Vector{ double x,y; Vector(double a=0,double b=0):x(a),y(b){} bool operator <(const Vector &a)const{ //return x<a.x||(x==a.x&&y<a.y); return sgn(x-a.x)<0||(sgn(x-a.x)==0&&sgn(y-a.y)<0); }};typedef Vector Point;Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}Vector operator *(Vector a,double b){return Vector(a.x*b,a.y*b);}Vector operator /(Vector a,double b){return Vector(a.x/b,a.y/b);}bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}int ConvexHull(Point p[],int n,Point ch[]){ sort(p+1,p+1+n); int m=0; for(int i=1;i<=n;i++){ while(m>1&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<0) m--; ch[++m]=p[i]; } int k=m; for(int i=n-1;i>=1;i--){ while(m>k&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<0) m--; ch[++m]=p[i]; } if(n>1) m--; return m;}int n,r,x,y;double ans;Point p[N],ch[N];bool solve(){ for(int i=2;i<=n;i++) if(sgn(Cross(ch[i]-ch[i-1],ch[i%n+1]-ch[i]))!=0 &&sgn(Cross(ch[i%n+1]-ch[i],ch[(i+1)%n+1]-ch[i%n+1]))!=0) return false; return true;}int main(int argc, const char * argv[]) { int T=read(); while(T--){ n=read(); for(int i=1;i<=n;i++) p[i].x=read(),p[i].y=read(); if(n<6) {puts("NO");continue;} n=ConvexHull(p,n,ch); if(solve()) puts("YES");else puts("NO"); } return 0;}
POJ 1228 Grandpa's Estate [稳定凸包]
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。