码迷,mamicode.com
首页 > 其他好文 > 详细

BNU 4067 求圆并

时间:2014-08-20 15:59:42      阅读:199      评论:0      收藏:0      [点我收藏+]

标签:style   blog   color   os   io   数据   for   ar   

好久没写过单组数据的题目了 QAQ

赤裸裸的模板题

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define sqr(x) ((x) * (x))

const int MAXN = 55;
const double EPS = 1e-8;
const double PI = acos(-1.0);//3.14159265358979323846
const double INF = 1;
const double eps = 1e-8;

int sgn(double x){
    if(x > eps)    return 1;
    else if(x < - eps)  return -1;
    else    return 0;
}
struct POINT {
    double x, y, ag;
    POINT() {}
    POINT(double x, double y): x(x), y(y) {}
    void read() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator == (const POINT &rhs) const {
        return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
    }
    bool operator < (const POINT &rhs) const {
        if(y != rhs.y) return y < rhs.y;
        return x < rhs.x;
    }
    POINT operator + (const POINT &rhs) const {
        return POINT(x + rhs.x, y + rhs.y);
    }
    POINT operator - (const POINT &rhs) const {
        return POINT(x - rhs.x, y - rhs.y);
    }
    POINT operator * (const double &b) const {
        return POINT(x * b, y * b);
    }
    POINT operator / (const double &b) const {
        return POINT(x / b, y / b);
    }
    double operator * (const POINT &rhs) const {
        return x * rhs.x + y * rhs.y;
    }
    double length() {
        return sqrt(x * x + y * y);
    }
    double angle() {
        return atan2(y, x);
    }
    POINT unit() {
        return *this / length();
    }
    void makeAg() {
        ag = atan2(y, x);
    }
    void print() {
        printf("%.10f %.10f\n", x, y);
    }
};
typedef POINT Vector;

double dist(const POINT &a, const POINT &b) {
    return (a - b).length();
}

double cross(const POINT &a, const POINT &b) {
    return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn right
double cross(const POINT &sp, const POINT &ed, const POINT &op) {
    return cross(sp - op, ed - op);
}

double area(const POINT& a, const POINT &b, const POINT &c) {
    return fabs(cross(a - c, b - c)) / 2;
}
//counter-clockwise
POINT rotate(const POINT &p, double angle, const POINT &o = POINT(0, 0)) {
    POINT t = p - o;
    double x = t.x * cos(angle) - t.y * sin(angle);
    double y = t.y * cos(angle) + t.x * sin(angle);
    return POINT(x, y) + o;
}

double cosIncludeAngle(const POINT &a, const POINT &b, const POINT &o) {
    POINT p1 = a - o, p2 = b - o;
    return (p1 * p2) / (p1.length() * p2.length());
}

double includedAngle(const POINT &a, const POINT &b, const POINT &o) {
    return acos(cosIncludeAngle(a, b, o));
    /*
    double ret = abs((a - o).angle() - (b - o).angle());
    if(sgn(ret - PI) > 0) ret = 2 * PI - ret;
    return ret;
    */
}

struct SEG {
    POINT st, ed;
    double ag;
    SEG() {}
    SEG(POINT st, POINT ed): st(st), ed(ed) {}
    void read() {
        st.read(); ed.read();
    }
    void makeAg() {
        ag = atan2(ed.y - st.y, ed.x - st.x);
    }
};
typedef SEG Line;

//ax + by + c > 0
Line buildLine(double a, double b, double c) {
    if(sgn(a) == 0 && sgn(b) == 0) return Line(POINT(sgn(c) > 0 ? -1 : 1, INF), POINT(0, INF));
    if(sgn(a) == 0) return Line(POINT(sgn(b), -c/b), POINT(0, -c/b));
    if(sgn(b) == 0) return Line(POINT(-c/a, 0), POINT(-c/a, sgn(a)));
    if(b < 0) return Line(POINT(0, -c/b), POINT(1, -(a + c) / b));
    else return Line(POINT(1, -(a + c) / b), POINT(0, -c/b));
}

void moveRight(Line &v, double r) {
    double dx = v.ed.x - v.st.x, dy = v.ed.y - v.st.y;
    dx = dx / dist(v.st, v.ed) * r;
    dy = dy / dist(v.st, v.ed) * r;
    v.st.x += dy; v.ed.x += dy;
    v.st.y -= dx; v.ed.y -= dx;
}

bool isOnSEG(const SEG &s, const POINT &p) {
    return (p == s.st || p == s.ed) ||
        (((p.x - s.st.x) * (p.x - s.ed.x) < 0 ||
          (p.y - s.st.y) * (p.y - s.ed.y) < 0) &&
         sgn(cross(s.ed, p, s.st)) == 0);
}

bool isInSEGRec(const SEG &s, const POINT &p) {
    return sgn(min(s.st.x, s.ed.x) - p.x) <= 0 && sgn(p.x - max(s.st.x, s.ed.x)) <= 0
        && sgn(min(s.st.y, s.ed.y) - p.y) <= 0 && sgn(p.y - max(s.st.y, s.ed.y)) <= 0;
}

bool isIntersected(const POINT &s1, const POINT &e1, const POINT &s2, const POINT &e2) {
    return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
        (max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
        (max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
        (max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
        (cross(s2, e1, s1) * cross(e1, e2, s1) >= 0) &&
        (cross(s1, e2, s2) * cross(e2, e1, s2) >= 0);
}

bool isIntersected(const SEG &a, const SEG &b) {
    return isIntersected(a.st, a.ed, b.st, b.ed);
}

bool isParallel(const SEG &a, const SEG &b) {
    return sgn(cross(a.ed - a.st, b.ed - b.st)) == 0;
}

//return Ax + By + C =0 ‘s A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
    A = L.ed.y - L.st.y;
    B = L.st.x - L.ed.x;
    C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}
//POINT of intersection
POINT operator * (const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    POINT I;
    I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
    I.y =   (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
    return I;
}

bool isEqual(const Line &a, const Line &b) {
    double A1, B1, C1;
    double A2, B2, C2;
    Coefficient(a, A1, B1, C1);
    Coefficient(b, A2, B2, C2);
    return sgn(A1 * B2 - A2 * B1) == 0 && sgn(A1 * C2 - A2 * C1) == 0 && sgn(B1 * C2 - B2 * C1) == 0;
}

double POINT_to_Line(const POINT &p, const Line &L) {
    return fabs(cross(p, L.st, L.ed)/dist(L.st, L.ed));
}

double POINT_to_SEG(const POINT &p, const SEG &L) {
    if(sgn((L.ed - L.st) * (p - L.st)) < 0) return dist(p, L.st);
    if(sgn((L.st - L.ed) * (p - L.ed)) < 0) return dist(p, L.ed);
    return POINT_to_Line(p, L);
}

double SEG_to_SEG(const SEG &a, const SEG &b) {
    double ans1 = min(POINT_to_SEG(a.st, b), POINT_to_SEG(a.ed, b));
    double ans2 = min(POINT_to_SEG(b.st, a), POINT_to_SEG(b.ed, a));
    return min(ans1, ans2);
}

struct Circle {
    POINT c;
    double r;
    Circle() {}
    Circle(POINT c, double r): c(c), r(r) {}
    void read() {
        c.read();
        scanf("%lf", &r);
    }
    double area() const {
        return PI * r * r;
    }
    bool contain(const Circle &rhs) const {
        return sgn(dist(c, rhs.c) + rhs.r - r) <= 0;
    }
    bool contain(const POINT &p) const {
        return sgn(dist(c, p) - r) <= 0;
    }
    bool intersect(const Circle &rhs) const {
        return sgn(dist(c, rhs.c) - r - rhs.r) < 0;
    }
    bool tangency(const Circle &rhs) const {
        return sgn(dist(c, rhs.c) - r - rhs.r) == 0;
    }
    POINT pos(double angle) const {
        POINT p = POINT(c.x + r, c.y);
        return rotate(p, angle, c);
    }
};

double CommonArea(const Circle &A, const Circle &B) {
    double area = 0.0;
    const Circle & M = (A.r > B.r) ? A : B;
    const Circle & N = (A.r > B.r) ? B : A;
    double D = dist(M.c, N.c);
    if((D < M.r + N.r) && (D > M.r - N.r)) {
        double cosM = (M.r * M.r + D * D - N.r * N.r) / (2.0 * M.r * D);
        double cosN = (N.r * N.r + D * D - M.r * M.r) / (2.0 * N.r * D);
        double alpha = 2 * acos(cosM);
        double beta = 2 * acos(cosN);
        double TM = 0.5 * M.r * M.r * (alpha - sin(alpha));
        double TN = 0.5 * N.r * N.r * (beta - sin(beta));
        area = TM + TN;
    }
    else if(D <= M.r - N.r) {
        area = N.area();
    }
    return area;
}

int intersection(const SEG &s, const Circle &cir, POINT &p1, POINT &p2) {
    double angle = cosIncludeAngle(s.ed, cir.c, s.st);
    //double angle1 = cos(includedAngle(s.ed, cir.c, s.st));
    double B = dist(cir.c, s.st);
    double a = 1, b = -2 * B * angle, c = sqr(B) - sqr(cir.r);
    double delta = sqr(b) - 4 * a * c;
    if(sgn(delta) < 0) return 0;
    if(sgn(delta) == 0) delta = 0;
    double x1 = (-b - sqrt(delta)) / (2 * a), x2 = (-b + sqrt(delta)) / (2 * a);
    Vector v = (s.ed - s.st).unit();
    p1 = s.st + v * x1;
    p2 = s.st + v * x2;
    return 1 + sgn(delta);
}

double CommonArea(const Circle &cir, POINT p1, POINT p2) {
    if(p1 == cir.c || p2 == cir.c) return 0;
    if(cir.contain(p1) && cir.contain(p2)) {
        return area(cir.c, p1, p2);
    } else if(!cir.contain(p1) && !cir.contain(p2)) {
        POINT q1, q2;
        int t = intersection(Line(p1, p2), cir, q1, q2);
        if(t == 0) {
            double angle = includedAngle(p1, p2, cir.c);
            return 0.5 * sqr(cir.r) * angle;
        } else {
            double angle1 = includedAngle(p1, p2, cir.c);
            double angle2 = includedAngle(q1, q2, cir.c);
            if(isInSEGRec(SEG(p1, p2), q1))return 0.5 * sqr(cir.r) * (angle1 - angle2 + sin(angle2));
            else return 0.5 * sqr(cir.r) * angle1;
        }
    } else {
        if(cir.contain(p2)) swap(p1, p2);
        POINT q1, q2;
        intersection(Line(p1, p2), cir, q1, q2);
        double angle = includedAngle(q2, p2, cir.c);
        double a = area(cir.c, p1, q2);
        double b = 0.5 * sqr(cir.r) * angle;
        return a + b;
    }
}

struct Triangle {
    POINT p[3];
    Triangle() {}
    Triangle(POINT *t) {
        for(int i = 0; i < 3; ++i) p[i] = t[i];
    }
    void read() {
        for(int i = 0; i < 3; ++i) p[i].read();
    }
    double area() const {
        return ::area(p[0], p[1], p[2]);
    }
    POINT& operator[] (int i) {
        return p[i];
    }
};

double CommonArea(Triangle tir, const Circle &cir) {
    double ret = 0;
    ret += sgn(cross(tir[0], cir.c, tir[1])) * CommonArea(cir, tir[0], tir[1]);
    ret += sgn(cross(tir[1], cir.c, tir[2])) * CommonArea(cir, tir[1], tir[2]);
    ret += sgn(cross(tir[2], cir.c, tir[0])) * CommonArea(cir, tir[2], tir[0]);
    return abs(ret);
}

struct POLY {
    int n;
    POINT p[MAXN];//p[n] = p[0]
    void init(POINT *pp, int nn) {
        n = nn;
        for(int i = 0; i < n; ++i) p[i] = pp[i];
        p[n] = p[0];
    }
    double area() {
        if(n < 3) return 0;
        double s = p[0].y * (p[n - 1].x - p[1].x);
        for(int i = 1; i < n; ++i)
            s += p[i].y * (p[i - 1].x - p[i + 1].x);
        return s / 2;
    }
};
//the convex hull is clockwise
void Graham_scan(POINT *p, int n, int *stk, int &top) {//stk[0] = stk[top]
    sort(p, p + n);
    top = 1;
    stk[0] = 0; stk[1] = 1;
    for(int i = 2; i < n; ++i) {
        while(top && cross(p[i], p[stk[top]], p[stk[top - 1]]) <= 0) --top;
        stk[++top] = i;
    }
    int len = top;
    stk[++top] = n - 2;
    for(int i = n - 3; i >= 0; --i) {
        while(top != len && cross(p[i], p[stk[top]], p[stk[top - 1]]) <= 0) --top;
        stk[++top] = i;
    }
}
//use for half_planes_cross
bool cmpAg(const Line &a, const Line &b) {
    if(sgn(a.ag - b.ag) == 0)
        return sgn(cross(b.ed, a.st, b.st)) < 0;
    return a.ag < b.ag;
}
//clockwise, plane is on the right
bool half_planes_cross(Line *v, int vn, POLY &res, Line *deq) {
    int i, n;
    sort(v, v + vn, cmpAg);
    for(i = n = 1; i < vn; ++i) {
        if(sgn(v[i].ag - v[i-1].ag) == 0) continue;
        v[n++] = v[i];
    }
    int head = 0, tail = 1;
    deq[0] = v[0], deq[1] = v[1];
    for(i = 2; i < n; ++i) {
        if(isParallel(deq[tail - 1], deq[tail]) || isParallel(deq[head], deq[head + 1]))
            return false;
        while(head < tail && sgn(cross(v[i].ed, deq[tail - 1] * deq[tail], v[i].st)) > 0)
            --tail;
        while(head < tail && sgn(cross(v[i].ed, deq[head] * deq[head + 1], v[i].st)) > 0)
            ++head;
        deq[++tail] = v[i];
    }
    while(head < tail && sgn(cross(deq[head].ed, deq[tail - 1] * deq[tail], deq[head].st)) > 0)
        --tail;
    while(head < tail && sgn(cross(deq[tail].ed, deq[head] * deq[head + 1], deq[tail].st)) > 0)
        ++head;
    if(tail <= head + 1) return false;
    res.n = 0;
    for(i = head; i < tail; ++i)
        res.p[res.n++] = deq[i] * deq[i + 1];
    res.p[res.n++] = deq[head] * deq[tail];
    res.n = unique(res.p, res.p + res.n) - res.p;
    res.p[res.n] = res.p[0];
    return true;
}

//ix and jx is the POINTs whose distance is return, res.p[n - 1] = res.p[0], res must be clockwise
double dia_rotating_calipers(POLY &res, int &ix, int &jx) {
    double dia = 0;
    int q = 1;
    for(int i = 0; i < res.n - 1; ++i) {
        while(sgn(cross(res.p[i], res.p[q + 1], res.p[i + 1]) - cross(res.p[i], res.p[q], res.p[i + 1])) > 0)
            q = (q + 1) % (res.n - 1);
        if(sgn(dist(res.p[i], res.p[q]) - dia) > 0) {
            dia = dist(res.p[i], res.p[q]);
            ix = i; jx = q;
        }
        if(sgn(dist(res.p[i + 1], res.p[q]) - dia) > 0) {
            dia = dist(res.p[i + 1], res.p[q]);
            ix = i + 1; jx = q;
        }
    }
    return dia;
}
//a and b must be clockwise, find the minimum distance between two convex hull
double half_rotating_calipers(POLY &a, POLY &b) {
    int sa = 0, sb = 0;
    for(int i = 0; i < a.n; ++i) if(sgn(a.p[i].y - a.p[sa].y) < 0) sa = i;
    for(int i = 0; i < b.n; ++i) if(sgn(b.p[i].y - b.p[sb].y) < 0) sb = i;
    double tmp, ans = dist(a.p[0], b.p[0]);
    for(int i = 0; i < a.n; ++i) {
        while(sgn(tmp = cross(a.p[sa], a.p[sa + 1], b.p[sb + 1]) - cross(a.p[sa], a.p[sa + 1], b.p[sb])) > 0)
            sb = (sb + 1) % (b.n - 1);
        if(sgn(tmp) < 0) ans = min(ans, POINT_to_SEG(b.p[sb], SEG(a.p[sa], a.p[sa + 1])));
        else ans = min(ans, SEG_to_SEG(SEG(a.p[sa], a.p[sa + 1]), SEG(b.p[sb], b.p[sb + 1])));
        sa = (sa + 1) % (a.n - 1);
    }
    return ans;
}

double rotating_calipers(POLY &a, POLY &b) {
    return min(half_rotating_calipers(a, b), half_rotating_calipers(b, a));
}

/*******************************************************************************************/

POINT p[MAXN];
Circle cir1, cir2, cir3;

int main(){
    double x1, y1, R, r, x2, y2, k;
    scanf("%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&R,&r,&x2,&y2,&k);
    cir1 = Circle(POINT(x1, y1), R);
    cir2 = Circle(POINT(x1, y1), r);
    cir3 = Circle(POINT(x2, y2), k);
    printf("%.2f\n",CommonArea(cir3, cir1) - CommonArea(cir3, cir2));
}

 

BNU 4067 求圆并,布布扣,bubuko.com

BNU 4067 求圆并

标签:style   blog   color   os   io   数据   for   ar   

原文地址:http://www.cnblogs.com/wushuaiyi/p/3924562.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!