POJ 2079 Triangle (凸包+旋转卡壳，求最大三角形面积)

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 8406		Accepted: 2490

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

给出n个点,求可以构成的最大的三角形面积。

思路：三角形的点挑选肯定在凸包上,但一一枚举O(n^3)会超时,这里就用到了旋转卡(qia)壳,也是第一次接触,计算几何为了速学基本也是参考大牛kuangbin代码

这里是旋转卡(qia)壳的资料链接：

http://blog.csdn.net/ACMaker
http://www.cnblogs.com/xdruid/archive/2012/07/01/2572303.html

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
using namespace std;

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)
{
    return (a-b)*(a-b);
}
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)
{
    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;
    }
}
//旋转卡壳，求两点间距离平方的最大值进行了修改可求最大三角面积
int rotating_calipers(Point p[],int n)
{
    int ans = 0;
    for(int i = 0;i < n;i++)
    {
        int j = (i+1)%n;
        int k = (j+1)%n;
        while(j != i && k != i)
        {
            ans = max(ans,abs((p[j]-p[i])^(p[k]-p[i])) );
            while( ( (p[i]-p[j])^(p[(k+1)%n]-p[k]) ) < 0 )
                k = (k+1)%n;
            j = (j+1)%n;
        }
    }
    return ans;
}
Point p[MAXN];
int main()
{
    int n;
    while(scanf("%d",&n) == 1)
    {
        if(n == -1)break;
        for(int i = 0;i < n;i++)
            list[i].input();
        Graham(n);
        for(int i = 0;i < top;i++)
            p[i] = list[Stack[i]];
        int ans = rotating_calipers(p,top);
        printf("%.2lf\n",ans/2.0);
    }
    return 0;
}