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POJ 2079 凸包最大內接三角形

編輯：C++入門知識

Triangle Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 8038 Accepted: 2375

Description

Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points.

Input

The input consists of several test cases. The first line of each test case contains an integer n, indicating the number of points on the plane. Each of the following n lines contains two integer xi and yi, indicating the ith points. The last line of the input is an integer ?1, indicating the end of input, which should not be processed. You may assume that 1 <= n <= 50000 and ?10⁴ <= xi, yi <= 10⁴ for all i = 1 . . . n.

Output

For each test case, print a line containing the maximum area, which contains two digits after the decimal point. You may assume that there is always an answer which is greater than zero.

Sample Input

Sample Output

0.50
27.00

經典題目：

代碼：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/10 16:26:51
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-5
#define pi acos(-1.0)
typedef long long ll;
int dcmp(double x){
	if(fabs(x)0?1:-1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0){
		x=_x;y=_y;
	}
};
Point operator + (const Point &a,const Point &b){  
    return Point(a.x+b.x,a.y+b.y);  
}  
Point operator - (const Point &a,const Point &b){  
    return Point(a.x-b.x,a.y-b.y);  
}  
Point operator * (const Point &a,const double &p){  
    return Point(a.x*p,a.y*p);  
}  
Point operator / (const Point &a,const double &p){  
    return Point(a.x/p,a.y/p);  
}  
bool operator < (const Point &a,const Point &b){  
    return a.x 0) return Length(v3);  
    else return fabs(Cross(v1, v2)) / Length(v1);  
}  
double dis_pair_seg(Point p1, Point p2, Point p3, Point p4)  {  
    return min(min(DistanceToSegment(p1, p3, p4), DistanceToSegment(p2, p3, p4)),  
     min(DistanceToSegment(p3, p1, p2), DistanceToSegment(p4, p1, p2)));  
}  
vector CH(vector p){
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()),p.end());
	int n=p.size();
	int m=0;
	vector ch(n+1);
	for(int i=0;i1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1])<=0)m--;
		ch[m++]=p[i];
	}
	int k=m;
	for(int i=n-2;i>=0;i--){
		while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
		ch[m++]=p[i];
	}
	if(n>1)m--;
	ch.resize(m);
	return ch;
}
double RC_Distance(vector ch1,vector ch2)  
{  
    int q=0, p=0,n=ch1.size(),m=ch2.size();
    for(int i=0;i eps) q=i;  
    ch1.push_back(ch1[0]);ch2.push_back(ch2[0]); 
  
    double tmp, ans=1e100;  
    for(int i=0;i eps)  
            q=(q+1)%m;  
        if(tmp < -eps) ans = min(ans,DistanceToSegment(ch2[q],ch1[p],ch1[p+1]));  
        else ans = min(ans,dis_pair_seg(ch1[p],ch1[p+1],ch2[q],ch2[q+1]));  
        p=(p+1)%n;  
    }  
    return ans;  
}  
double RC_Triangle(vector res)// 凸包最大內接三角形  
{  
	 int n=res.size();
     if(n<3)    return 0;  
     double ans=0, tmp;  
     res.push_back(res[0]);
     int j, k;  
     for(int i=0;i Cross(res[j] - res[i], res[k] - res[i])) k= (k+1)%n;  
              tmp = Cross(res[j] - res[i], res[k] - res[i]);if(tmp > ans) ans = tmp;  
              j = (j+1)%n;  
         }  
     }  
     return ans/2;  
}  

int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,m;
	 while(cin>>n&&n!=-1){
		 vector res;
		 Point p;
		 while(n--)scanf("%lf%lf",&p.x,&p.y),res.push_back(p);
		 res=CH(res);
		 printf("%.2lf\n",RC_Triangle(res));
	 }
     return 0;
}