poj 2079 Triangle(旋转卡壳)

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 7637		Accepted: 2238

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

Source

Shanghai 2004 Preliminary

 

 

 

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;
#define eps 1e-8

struct Point
{
    int x,y;
    Point(int _x = 0, int _y = 0)
    {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x, y - b.y);
    }
    int operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    int operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    void input()
    {
        scanf("%d%d",&x,&y);
    }
};
int dist2(Point a,Point b)//距离的平方！！！attention！！ 
{
    return (a-b)*(a-b);
}


int dcmp(double a)
{
    if(fabs(a)<eps)return 0;
    if(a>0)return 1;
    else return -1; 
}

const int MAXN = 50010;
Point list[MAXN];
int Stack[MAXN],top;
bool _cmp(Point p1,Point p2)
{
    int tmp = (p1-list[0])^(p2-list[0]);
    if(tmp > 0)return true;
    else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
        return true;
    else return false;
}

void Graham(int n)//求凸包 
{
    Point p0;
    int k = 0;
    p0 = list[0];
    for(int i = 1;i < n;i++)
        if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
        {
            p0 = list[i];
            k = i;
        }
    swap(list[0],list[k]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0;
        Stack[1] = 1;
        return;
    }
    Stack[0] = 0;
    Stack[1] = 1;
    top = 2;
    for(int i = 2;i < n;i++)
    {
        while(top > 1 && ((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0 )
            top--;
        Stack[top++] = i;
    }
}

double Area(Point *p,int n)
{
    double area=0;
    for(int i=0; i<n; i++){
        area+=(p[i]^p[(i+1)%n]);
    }
    return fabs((double)(area)/2);
}

double rotating_calipers(Point p[],int n)
{
    double ans = 0;
    Point v;
    int cur = 1;
    for(int i = 0;i < n;i++)//先转k再转j再转i!!! 
    {
        v = p[i]-p[(i+1)%n];
        int k,j;
        j=(i+1)%n;
        k=(j+1)%n;
        while(k!=i&&j!=i)
        {      
            v = p[i]-p[j];//wa无数次！忘记了j也在转啊！！！    
               while(dcmp(v^(p[(k+1)%n]-p[k])) < 0)    
            {   
                k = (k+1)%n;            
            }
            Point aa[4];
            aa[0]=p[i];aa[1]=p[j%n];aa[2]=p[k%n];
            double tmp=Area(aa,3);  
            ans=max(ans,tmp);
            j=(j+1)%n;
        }
           Point aa[4];
        aa[0]=p[i];aa[1]=p[j%n];aa[2]=p[k%n];
        double tmp=Area(aa,3);  
        ans=max(ans,tmp);
    }
    return ans;
}
Point p[MAXN];
 
int main()
{
    int n;
    while(scanf("%d",&n) == 1&&n!=-1)
    {
        for(int i = 0;i < n;i++)
            list[i].input();
        Graham(n);
        for(int i = 0;i < top;i++)
        {
            p[i] = list[Stack[i]];
        }
        printf("%.2f
",rotating_calipers(p,top)) ;
    }
    return 0;
}