HDU 6762 Mow (2020 Multi-University Training Contest 1 1012） 半平面交

Mow

题目链接

分析

将多边形的边向内部缩 r 个单位长度，然后这些边所围成的内部区域，就是圆心的合法范围，该范围也是一个多边形，假设面积是(a)，周长是(b)，那么可以知道圆可以覆盖的面积是 (a + b * r + pi *r^2)。现在问题转换为了求这些边所围成的区域，这正是半平面交所要做的事情。

需要用到的知识点：

1. 极角排序
2. 直线平移
3. 直线求交点
4. 单调队列求半平面交
5. 多边形利用三角剖分求面积，求周长

另外需要注意的是该题目精度有些卡，需要用longdouble，另外要注意一些常用函数后面加 l,比如acos要换成acosl，sqrt要换成sqrtl等。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
#define dbg(x...) do { cout << "33[32;1m" << #x <<" -> "; err(x); } while (0)
void err() { cout << "33[39;0m" << endl; }
template<class T, class... Ts> void err(const T& arg,const Ts&... args) { cout << arg << " "; err(args...); }
const int N = 200 + 5;

typedef long double db;
const db eps = 1e-12;
const db pi = acosl(-1.0);

int T, n;
db A, B, r;
int sgn(db x){
    if(fabs(x) < eps) return 0;
    if(x < 0) return -1;
    return 1;
}

inline db sqr(db x){return x * x;}
struct Point{
    db x, y;
    Point(){}
    Point(db x, db y):x(x), y(y){}
    void input(){
        cin >> x >> y;
    }
    bool operator == (Point b) const {
        return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
    }
    bool operator < (Point b) const{
        return sgn(x - b.x) == 0 ? sgn(y - b.y) < 0 : x < b.x;
    }
    // 向量运算
    Point operator - (const Point &b) const{
        return Point(x - b.x, y - b.y);
    }
    Point operator + (const Point &b) const{
        return Point(x + b.x, y + b.y);
    }
    db operator ^ (const Point &b) const{
        return x * b.y - y * b.x;
    }
    db operator * (const Point &b) const{
        return x * b.x + y * b.y;
    }
    // 标量运算 * / 
    Point operator *(const db &k) const{
        return Point(x * k , y * k);
    }
    Point operator / (const db &k) const{
        return Point(x / k, y / k);
    }

    db len(){
        return sqrtl(x * x + y * y);
    }
    db len2(){
        return x * x + y * y;
    }
    db distance(Point p) {
        return (*this - p).len();
    }

    Point trunc(db r) {
        db l = len();
        if(sgn(l) == 0) return *this;
        return Point(x * r / l, y * r / l);
    }
    Point rotleft(){
        return Point(-y, x);
    }
    Point rotright(){
        return Point(y, -x);
    }
}p[N];

struct Line{
    Point s, e;
    Line(){}
    Line(Point s, Point e) : s(s), e(e){}
    void input(){
        s.input();e.input();
    }
    bool operator == (Line v){
        return (s == v.s) && (e == v.e);
    }
    db length() {
        return s.distance(e);
    }
    bool parallel(Line v){
        return sgn((e - s) ^ (v.e - v.s)) == 0;
    }
    Point crossPoint(Line v) {
        db a1 = (v.e - v.s) ^ (s - v.s);
        db a2 = (v.e - v.s) ^ (e - v.s);
        db t = a2 - a1;
        return Point((s.x * a2 - e.x * a1) / t, (s.y * a2 - e.y * a1) / t);
    }
}l[N];
 
struct Polygon{
    int n;
    Point p[N];
    Line l[N];
    void input(int n) {
        this->n = n;
        for(int i=0;i<n;i++) p[i].input();
    }
    db getCircumference(){
        db res = 0;
        for(int i=0;i<n;i++) res += p[i].distance(p[(i+1)%n]);
        return res;
    }
    db getArea(){
        db res = 0;
        for(int i=0;i<n;i++)res += (p[i] ^ p[(i+1)%n]);
        return fabs(res) / 2; 
    }
    struct cmp{
        Point p;
        cmp(const Point &p0) {p = p0;}
        bool operator()(const Point &aa, const Point &bb) {
            Point a = aa, b = bb;
            int d = sgn((a - p) ^ (b - p));
            if(d == 0) return sgn(a.distance(p) - b.distance(p)) < 0;
            return d > 0;
        }
    };
    void norm(){
        Point mi = p[0]; for(int i=1;i<n;i++) mi = min(mi, p[i]);
        sort(p, p+n, cmp(mi));
    }
}lawn, poly;
void adjust(){
    lawn.norm();
    for(int i=0;i<n;i++){
        Point a = lawn.p[i], b = lawn.p[(i+1)%n];
        Point aa = a + (b - a).rotleft().trunc(r);
        Point bb = b + (b - a).rotleft().trunc(r);
        l[i] = Line(aa, bb);
    }
}
bool getHalfPlanes(){
    Line q[N];
    Point t[N];
    int ll = 1, rr = 0;
    for(int i=0;i<n;i++){
        if(l[i] == l[(i-1+n)%n] || l[i].parallel(l[(i-1+n)%n]))continue;
        while(ll < rr && ((l[i].e - t[rr]) ^ (l[i].s - t[rr])) > eps) rr--;
        while(ll < rr && ((l[i].e - t[ll+1]) ^ (l[i].s - t[ll+1])) > eps) ll++;
        q[++rr] = l[i];
        if(ll < rr) t[rr] = q[rr].crossPoint(q[rr-1]);
    }
    while(ll < rr && ((q[ll].e - t[rr]) ^ (q[ll].s - t[rr])) > eps) rr--;
    t[rr+1] = q[ll].crossPoint(q[rr]);
    ++rr, ++ll;
    if(rr - ll + 1 <= 2) return false;
    poly.n = rr - ll + 1;
    for(int i=ll;i<=rr;i++) poly.p[i-ll] = t[i];
    return true;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("i.in","r",stdin);
//  freopen("o.out","w",stdout);
#endif
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        cin >> r >> A >> B;
        lawn.input(n);
        adjust();
        db total = lawn.getArea();
        db res = total * A;
        if(getHalfPlanes()) {
            db area = poly.getArea();
            db length = poly.getCircumference();
            area = area + length * r + pi * r * r;
            res = min(res, area * B + (total - area) * A);
        }
        cout << fixed << setprecision(20) << res << endl;
    }
    return 0;
}