题目链接：?id=2079
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 <= 104for 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
33 42 62 752 63 92 08 06 5-1
Sample Output
0.5027.00
Source
题意：
求最大三角形面积；
PS:
bin神模板！
代码如下：
#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,求三角形最大面积int rotating_calipers(Point p[],int n){int ans = 0;Point v;int cur = 1;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;}
，接受自己的失败面，是一种成熟，更是一种睿智
