uva10078The Art Gallery

最新推荐文章于 2021-06-26 10:18:29 发布

原创最新推荐文章于 2021-06-26 10:18:29 发布 · 538 阅读

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本文介绍了一种通过计算凸包来判断美术馆是否存在内部关键点的方法。首先找到美术馆边界上的点，然后利用Graham扫描算法确定凸包。如果凸包上的点数与总点数相等，则表明不存在关键点。

题意：判断美术馆是否有关键点，美术馆是一些点，这些点构成的最小边长的边界就是美术馆的边界，看看在这个边界中是否包含一些点，即有些点不在边界上，而在边界内部

解题：求出凸包，看凸包上的点个数是否等于美术馆所有点的个数，相等，没有关键点，小于，包含关键点。

代码：

//uva10078 The Art Gallery
//AC By Warteac
//Runtime:0.013s
//2013-5-14
#include<iostream>
#include<cstdio>
#include<vector>
#include<stack>
#include <iomanip>
#include<algorithm>
#include<cmath>
using namespace std;
typedef int PType;//定义点的类型
///////////////////////////////////////
struct point{
    PType x,y;
    point(PType a = 0, PType b = 0){x  = a; y = b;}
    void transfer(point p){x -= p.x, y -= p.y;}
    void restore(point p){x += p.x, y += p.y;}
    void print(){cout << "x = " << x << "y = " << y << endl;}  
};
int direction(point v1,point v2){
 	return (v1.x*v2.y-v1.y*v2.x);
 }
 bool cmp(point p1,point p2){
   return (p1.x*p2.y < p1.y*p2.x);
}
bool grham_scan(vector <point> p , vector <point> &parray){
//find first point p0 with minimum y-coordinate or minimun x-coordinate if y are the same
    PType x = p[0].x, y = p[0].y;
    int index = 0,i;
    for(i = 1; i < p.size(); i++){
        if(p[i].y > y) continue;
        if(p[i].y < y){
            y = p[i].y; x = p[i].x; index = i;
        }else if(p[i].x < x){
            y = p[i].y; x = p[i].x; index = i;
        }
    }
    point t = p[index];
    p[index] = p[0];
    p[0] = t;
//transfer points by p0
    for(int j = 0; j < p.size(); j++){
        p[j].transfer(t);
    }
//let (p1,p2……pm) sorted by polar angle
    sort(p.begin()+1,p.end(),cmp);
//delete those points with the same polar angle but the farthest ont from p0
    vector <point> ps;//saved in ps
    ps.clear();
    ps.push_back(p[0]);
    ps.push_back(p[1]);
    for(int i = 2; i < p.size(); i++){
        if(cmp(ps.back(),p[i])){
            ps.push_back(p[i]);   
        }else if(ps.back().x < p[i].x){
            ps.back() = p[i];
        }
    }
    if(ps.size() <= 2) return false;
//get the convex hull from ps
    parray.clear();
    parray.push_back(ps[0]);
    parray.push_back(ps[1]);
    ps.push_back(ps[0]);
    for(int i = 2; i < ps.size(); i++){
       point v1 (ps[i].x - parray.back().x, ps[i].y - parray.back().y);
       point v2 (parray.back().x - (parray.end()-2)->x, parray.back().y - (parray.end()-2)->y);
       int d = direction(v2,v1);
       if(d <= 0) {parray.push_back(ps[i]);}
       else if(d > 0){ 
       parray.pop_back();i--; 
       }
    }   
//restore the points' coordinate
    for(int j = 0; j < parray.size(); j++){
        parray[j].restore(t);
    }
    return true;
}
//////////////////////////////////////////////////////
int main(){
    int sites;
    PType x,y;
    vector <point> v;
    point t,mi;
    vector < vector <point> >  p;//convex hull
    int aflag[105] = {0};
    int area = 0;
    //input kingdom
    while(cin >> sites && sites){
        v.clear();
        vector <point> temp;//one convex hull
        while(sites--){
	        cin >> t.x >> t.y;
	        v.push_back(t);
        }
        //computing convex hull
	    if(grham_scan(v,temp) == true){
	        if(temp.size() == v.size()+1)
	            cout << "No" << endl;
	        else cout << "Yes" <<endl;
	    }else{
	        cout << "no convex hull" << endl;
	    }
     }
 return 0;
}