LightOJ 1203 Guarding Bananas (Convex Hull)

Jan's LightOJ :: Problem 1203 - Guarding Bananas

　　简单凸包。处理凸包的时候吧ch写成了pt，wa了两次。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
} ;

bool operator < (Point x, Point y) { return x.x < y.x || (x.x == y.x && x.y < y.y);}
bool operator == (Point x, Point y) { return x.x == y.x && x.y == y.y;}
Point operator - (Point x, Point y) { return Point(x.x - y.x, x.y - y.y);}

const double EPS = 1e-8;
inline int sgn(double x) { return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}
inline double crossDet(Point x, Point y) { return x.x * y.y - x.y * y.x;}
inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}

int andrew(Point *pt, int n, Point *ch) {
    sort(pt, pt + n);
    n = (int) (unique(pt, pt + n) - pt);
    int m = 0;
    for (int i = 0; i < n; i++) {
        while (m > 1 && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--;
        ch[m++] = pt[i];
    }
    int k = m;
    for (int i = n - 2; i >= 0; i--) {
        while (m > k && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--;
        ch[m++] = pt[i];
    }
    if (n > 1) m--;
    return m;
}

const int N = 111111;
Point pt[N], ch[N];
const double PI = acos(-1.0);

inline double angle(Point x) { return atan2(x.y, x.x);}

int main() {
//    freopen("in", "r", stdin);
    int T, n;
    cin >> T;
    for (int cas = 1; cas <= T; cas++) {
        cin >> n;
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &pt[i].x, &pt[i].y);
        }
        if (n <= 2) {
            printf("Case %d: 0.00000000\n", cas);
            continue;
        }
        n = andrew(pt, n, ch);
        for (int i = 0; i < 10; i++) {
            ch[i + n] = ch[i];
        }
        double mini = 1e10;
        for (int i = 0; i < n; i++) {
            double ang1 = angle(ch[i] - ch[i + 1]);
            double ang2 = angle(ch[i + 2] - ch[i + 1]);
            double da = fabs(ang1 - ang2);
            if (da > PI) da = 2.0 * PI - da;
            mini = min(mini, da * 180.0 / PI);
        }
        printf("Case %d: %.8f\n", cas, mini);
    }
    return 0;
}

——written by Lyon