UVALive 6489 Triangles LA 6489 Triangles

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=625&page=show_problem&problem=4500

题意：给N个点 求出组成的最小三角形面积和最大三角形面积  

详细题解：这丫的题目坑了我一个月，然后A了之后觉得自己的方法完全是个错的0.0

一开始的想法是：最大三角形面积很简单就是用凸包+旋转卡壳  ，最小三角形面积就是将每一个点极角排序。 然后求极点和每个相邻的两点组成的三角形的面积   之后发现这样有问题 如果这两边的夹角很大 但是长度很短  还是可能会面积最小 。 所以又想到以每个点为顶点，和其余点的用长度来排了下序，然后以长度的顺序来组成三角形之后发现这样测完所有数据要50s+

之后左搞右搞，发现原来极角排序写错了，干脆就删除了极角排序，然后发现还是超时30s+

然后无奈的乱改，把边排序的叉积的方法去掉，改为直接公式计算就9s了， 不曾想这竟然过了……

感觉完全就是逻辑错误，贴上代码，大家有什么正解的话，求在下面留言指教。。。。

#include<cstdio>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define N 2100#define inf 0x7fffffffconst double eps = 1e-8;int dcmp(double x){if(fabs(x) < eps) return 0;else return x < 0 ? -1 : 1;}struct Point{    double x, y;    Point(double x = 0, double y = 0) : x(x), y(y) {}};Point p[N];Point pp[N];Point ch[N];typedef Point Vector;Vector operator - (Point A, Point B) {return Vector(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;}bool operator != (const Point &a, const Point &b){return dcmp(a.x-b.x) != 0 || dcmp(a.y-b.y) != 0;}double Dot(Vector A, Vector B){return A.x*B.x + A.y*B.y;}double Length(Vector A){return sqrt(Dot(A, A));}double Cross(Vector A, Vector B){return A.x*B.y - A.y*B.x;}double Area2(Point A, Point B, Point C){return fabs(Cross(B-A, C-A));}Point tmp;bool cmp_bian(Point a, Point b){    double lena = (a.x-tmp.x)*(a.x-tmp.x)+(a.y-tmp.y)*(a.y-tmp.y);    double lenb = (b.x-tmp.x)*(b.x-tmp.x)+(b.y-tmp.y)*(b.y-tmp.y);    return dcmp(lena - lenb)< 0;    //return Length(a - tmp) < Length(b-tmp);}bool cmp ( Point a, Point b ){    if ( a.x != b.x ) return a.x < b.x;    else return a.y < b.y;}//凸包int ConvexHull(Point *p, int n, Point * ch){sort(p, p+n, cmp);int m = 0;for(int i = 0; i < n; i++){while(m > 1 && dcmp(Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) <= 0) m--;ch[m++] = p[i];}int k = m;for(int i = n-2; i >= 0; i--){while(m > k && dcmp(Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2])) <= 0) m--;ch[m++] = p[i];}if(n > 1) m--;return m;}//旋转卡壳求最大三角形面积double rotaing_calipers(Point ch[], int n){    int p;    int i, j;    double ans = 0;    for( i = 0; i < n-1; i++)    {        p = 1;        for( j = i+1; j < n; j++)        {            while(fabs(Cross(ch[j]-ch[i],ch[p+1]-ch[i])) > fabs((Cross(ch[j]-ch[i],ch[p]-ch[i]))))                p = (p+1) % (n-1);            ans = max(ans, fabs(Cross(ch[i]-ch[p],ch[j]-ch[p])));        }        ans = max(ans, fabs(Cross(ch[i]-ch[p],ch[j]-ch[p])));    }    return ans/2;}int main (){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n;    while(scanf("%d", &n),n)    {        for(int i = 0; i < n; i++)        {            scanf("%lf %lf", &p[i].x, &p[i].y);            pp[i].x = p[i].x;            pp[i].y = p[i].y;        }        double Min = inf, Max = -1;        for(int i = 0; i < n; i++)        {            tmp.x = p[i].x, tmp.y = p[i].y;            double temp ;            for(int j = 0; j < n; j++)            {                if(p[i] == pp[j]) continue;                if(pp[j+1] != p[i] && j+1 < n)                    temp = Area2(p[i], pp[j], pp[j+1])/2;                else if(pp[j+1] == p[i] && j+2 < n)                    temp = Area2(p[i], pp[j], pp[j+2])/2;                Min = min(Min, temp);                if(Min == 0) break;            }            sort(pp, pp+n, cmp_bian);            for(int j = 1; j < n-1; j++)            {                temp = Area2(p[i],pp[j],pp[j+1])/2;                Min = min(Min, temp);            }            if(Min == 0) break;        }        int len = ConvexHull(p, n, ch);        Max = rotaing_calipers(ch, len);        printf("%.1lf %.1lf\n", Min, Max);    }    return 0;}