首页 > 代码库 > POJ 2079 凸包最大内接三角形

POJ 2079 凸包最大内接三角形

2024-07-04 00:18:39   220人阅读

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

3
3 4
2 6
2 7
5
2 6
3 9
2 0
8 0
6 5
-1

Sample Output

0.50
27.00


经典题目：

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/10 16:26:51
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
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)<eps)return 0;
	return 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<b.x||(a.x==b.x&&a.y<b.y);  
}  
bool operator == (const Point &a,const Point &b){  
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;  
}  
double Dot(Point  a,Point b){  
    return a.x*b.x+a.y*b.y;  
}  
double Length(Point a){  
    return sqrt(Dot(a,a));  
}  
double Angle(Point a,Point b){  
    return acos(Dot(a,b)/Length(a)/Length(b));  
}  
double angle(Point a){  
    return atan2(a.y,a.x);  
}  
double Cross(Point a,Point b){  
    return a.x*b.y-a.y*b.x;  
}  
Point vecunit(Point a){  
    return a/Length(a);  
}  
Point Normal(Point a){  
    return Point(-a.y,a.x)/Length(a);  
}  
Point Rotate(Point a,double rad){  
    return Point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));  
}  
double Area2(Point a,Point b,Point c){  
    return Length(Cross(b-a,c-a));  
}  
double DistanceToSegment(Point p, Point a, Point b)  {  
    if(a == b) return Length(p-a);  
    Point v1 = b-a, v2 = p-a, v3 = p-b;  
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);  
    else if(dcmp(Dot(v1, v3)) > 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<Point> CH(vector<Point> p){
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()),p.end());
	int n=p.size();
	int m=0;
	vector<Point> ch(n+1);
	for(int i=0;i<n;i++){
		while(m>1&&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<Point> ch1,vector<Point> ch2)  
{  
    int q=0, p=0,n=ch1.size(),m=ch2.size();
    for(int i=0;i<n;i++) if(ch1[i].y-ch1[p].y < -eps) p=i;  
    for(int i=0;i<m;i++)if(ch2[i].y-ch2[q].y > eps) q=i;  
    ch1.push_back(ch1[0]);ch2.push_back(ch2[0]); 
  
    double tmp, ans=1e100;  
    for(int i=0;i<n;i++) 
    {  
        while((tmp = Cross(ch1[p+1]-ch1[p], ch2[q+1]-ch1[p]) - Cross(ch1[p+1]-ch1[p], ch2[q]- ch1[p])) > 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<Point> res)// 凸包最大内接三角形  
{  
	 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<n;i++)
     {  
         j = (i+1)%n;  
         k = (j+1)%n;  
         while((j != k) && (k != i))  
         {  
              while(Cross(res[j] - res[i], res[k+1] - res[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<Point> 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;
}



联系
我们