• 凸包模板


    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef long double ld;
    const ld eps = 1e-8;
    const int N = 50009;
    const ld pi = acos(-1);
    struct Point {
        ld x, y;
        Point(ld X = 0, ld Y = 0) { x = X, y = Y; }
        Point operator-(Point a) { return Point(x - a.x, y - a.y); }
        Point operator+(Point a) { return Point(x + a.x, y + a.y); }
        ld operator*(Point a) { return x * a.y - y * a.x; }
        ld operator^(Point a) { return x * a.x + y * a.y; }
        Point operator*(ld a) { return Point(x * a, y * a); }
        Point rotate(ld thi) {
            return Point(x * cos(thi) + y * sin(thi), -x * sin(thi) + y * cos(thi));
        }
        ld dis() { return sqrt(x * x + y * y); }
        void out() { printf("%.2Lf %.2Lf
    ", x, y); }
    } p[N];
    int stk[N], used[N];
    int dcmp(ld a, ld b) {
        if (fabs(a - b) < eps)
            return 0;
        else if (a > b)
            return 1;
        else
            return -1;
    }
    typedef Point Vector;
    struct Line {
        Point s;
        Vector dir;
    };
    ld angle(Vector v1, Vector v2) { return acos(v1 ^ v2) / v1.dis() / v2.dis(); }
    void out(ld x) { printf("%.2Lf
    ", x); }
    bool cmp(Point a, Point b ) { 
        if (dcmp(a.x, b.x)==0) { 
            return a.y > b.y;
        }
        return a.x < b.x;
    }
    void solve() {
        int n;
        cin >> n;
        ld a, b, r;
        cin >> a >> b >> r;
        int idn = 0;
        for (int i = 1; i <= n; i ++) {
            ld x, y, thi;
            cin >> x >> y >> thi;
            Point v1, v2, v3, v4;
            Point v = {x, y};
            v1 = {x + b/2 - r, y + a/2 - r};
            v2 = {x + b/2 - r, y - a/2 + r};
            v3 = {x - b/2 + r, y - a/2 + r};
            v4 = {x - b/2 + r, y + a/2 - r};
            v1 = v1 - v;
            v2 = v2 - v;
            v3 = v3 - v;
            v4 = v4 - v;
            v1 = v1.rotate(-thi);
            v2 = v2.rotate(-thi);
            v3 = v3.rotate(-thi);
            v4 = v4.rotate(-thi);
            p[++idn] = v1 - v;
            p[++idn] = v2 - v;
            p[++idn] = v3 - v;
            p[++idn] = v4 - v;
        }
        n = idn;
        sort(p + 1, p + 1 + n, cmp);
        int tp = 1;
        stk[tp] = 1;
        for (int i = 2; i <= n; i ++) {
            while (tp >= 2 && dcmp((p[i] - p[stk[tp]]) * (p[i] - p[stk[tp-1]]), 0) >= 0) { 
                used[stk[tp--]] = 0;
            }
            stk[++tp] = i;
            used[i] = 1;
        }
        int tmp = tp;
        for (int i = n-1; i >= 1; i--) {
            if (used[i])continue;
            while (tp >= tmp + 1 && dcmp((p[i] - p[stk[tp]]) * (p[i] - p[stk[tp-1]]), 0)  >= 0)tp--;
            stk[++tp] = i;
            used[i] = 1;
        }
        ld ans = 0;
        for (int i = 1; i < tp; i ++) {
            ans += (p[stk[i + 1]] - p[stk[i]]).dis();
        }
        printf("%.2Lf
    ", ans + 2 * pi * r);
    }
    signed main() {
        ll t = 1;//cin >> t;
        while (t--) {
           solve();
        }
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Xiao-yan/p/15423724.html
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