LightOj1203 - Guarding Bananas（凸包求多边形中的最小角）

题目链接：http://lightoj.com/volume_showproblem.php?problem=1203

题意：给你一个点集，求凸包中最小的角；模板题，但是刚开始的时候模板带错了，错的我都想吐了；

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
const double eps = 1e-10;
const double PI = acos(-1);
const int N = 150010;

struct point
{
    double x, y;
    point(double x=0, double y=0) : x(x), y(y){}
    friend point operator - (const point& p1, const point& p2)
    {
        return point(p1.x-p2.x, p1.y-p2.y);
    }
    friend double operator ^ (const point& p1, const point& p2)
    {
        return p1.x*p2.y - p1.y*p2.x;
    }
}p[N], res[N];
double Dist(point p1, point p2)
{
    double dx = p1.x - p2.x, dy = p1.y - p2.y;
    return sqrt(dx*dx + dy*dy);
}
bool cmp1(point p1, point p2)
{
    if(p1.y == p2.y)
        return p1.x < p2.x;
    return p1.y < p2.y;
}
bool cmp2(point p1, point p2)///极角排序；若极角相同，距离近的在前面；
{
    double k = (p1-p[0])^(p2-p[0]);
    if( k>eps || (fabs(k)<eps && Dist(p1, p[0]) < Dist(p2, p[0]) ))
        return 1;
    return 0;
}
int Graham(int n)
{
    res[0] = p[0];if(n == 1) return 1;
    res[1] = p[1];if(n == 2) return 2;
    int top = 1;
    for(int i=2; i<n; i++)
    {
        while(top && ((res[top]-res[top-1])^(p[i]-res[top-1])) <= 0) top--;
        res[++top] = p[i];
    }
    return top+1;
}
int main()
{
    int n, T, tCase = 1;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i=0; i<n; i++)
            scanf("%lf %lf", &p[i].x, &p[i].y);
        sort(p, p+n, cmp1);///p[0]为最下方靠左的点；
        sort(p+1, p+n, cmp2);///以p[0]为基点，按叉积进行排序；
        int cnt = Graham(n);///求凸包的顶点个数cnt，保存在res中，下标从0开始；
        if(cnt < 3)
        {
            printf("Case %d: 0\n", tCase++);
            continue;
        }
        double ans = 1000;
        res[cnt] = res[0], res[cnt+1] = res[1];
        for(int i=1; i<=cnt; i++)
        {
            double a = Dist(res[i-1], res[i+1]);
            double b = Dist(res[i-1], res[i]);
            double c = Dist(res[i], res[i+1]);
            ans = min(ans, acos((b*b+c*c-a*a)/(2*b*c)));
        }
        printf("Case %d: %.6f\n", tCase++, ans*180/PI);
    }
    return 0;
}