Lightoj1203 Guarding Bananas(凸包)

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Lightoj1203

题目

有一片香蕉园，一只猴子想知道能看到所有香蕉的最小角度。

思路

求凸包，枚举角度。

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define pi acos (-1)#define rotate Rotateusing namespace std;typedef long long ll;const double eps = 1e-8;const int maxn=100010;const double INF = 1e20;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 operator - (point a) const    {        return point (x-a.x, y-a.y);    }    point operator + (point a) const    {        return point (x+a.x, y+a.y);    }    bool operator < (const point &a) const    {        return x < a.x || (x == a.x && y < a.y);    }    bool operator == (const point &a) const    {        return dcmp (x-a.x) == 0 && dcmp (y-a.y) == 0;    }};point operator * (point a, double p){    return point (a.x*p, a.y*p);}double cross (point a, point b){    return a.x*b.y-a.y*b.x;}double dot (point a, point b){    return a.x*b.x + a.y*b.y;}int ConvexHull (point *p, point *ch, int n){    sort (p, p+n);    int m = 0;    for (int i = 0; i < n; i++)    {        while (m > 1 && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    int k = m;    for (int i = n-2; i >= 0; i--)    {        while (m > k && cross (ch[m-1]-ch[m-2], p[i]-ch[m-1]) <= 0)            m--;        ch[m++] = p[i];    }    if (n > 1)        m--;    return m;}int cnt,n;point p[maxn],ch[maxn];double rad (point a, point b)  //两个向量的夹角{    return fabs (atan2 (fabs (cross (a, b)), dot (a, b)) );}double rad_degree (double rad)  //弧度转化为角度{    return rad/pi*180;}int main(){    int T;    scanf("%d",&T);    int kase=1;    while(T--)    {        printf ("Case %d: ", kase++);        scanf("%d",&n);        for(int i=0; i<n; i++)            scanf("%lf %lf",&p[i].x,&p[i].y);        double ans=INF;        cnt=ConvexHull(p,ch,n);        for(int i=0;i<cnt;i++)        {            ans=min(ans,rad(ch[(i+1)%cnt]-ch[i],ch[(i-1+cnt)%cnt]-ch[i]));        }        printf ("%.10f\n", rad_degree(ans));    }    return 0;}

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