圆的k次面积并

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//china no.1

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <vector>

#include <iostream>

#include <string>

#include <map>

#include <stack>

#include <cstring>

#include <queue>

#include <list>

#include <stdio.h>

#include <set>

#include <algorithm>

#include <cstdlib>

#include <cmath>

#include <iomanip>

#include <cctype>

#include <sstream>

#include <functional>

#include <stdlib.h>

#include <time.h>

#include <bitset>

using namespace std;

#define pi acos(-1)

#define PI acos(-1)

#define endl '\n'

#define srand() srand(time(0));

#define me(x,y) memset(x,y,sizeof(x));

#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)

#define close() ios::sync_with_stdio(0); cin.tie(0);

#define FOR(x,n,i) for(int i=x;i<=n;i++)

#define FOr(x,n,i) for(int i=x;i<n;i++)

#define W while

#define sgn(x) ((x) < 0 ? -1 : (x) > 0)

#define bug printf("***********\n");

#define db double

typedef long long LL;

const int INF=0x3f3f3f3f;

const LL LINF=0x3f3f3f3f3f3f3f3fLL;

const int dx[]={-,,,,,-,-,};

const int dy[]={,,,-,-,,-,};

const int maxn=1e3+;

const int maxx=1e6+;

const double EPS=1e-;

const double eps=1e-;

const int mod=;

template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}

template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}

template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}

template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}

template <class T>

inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return ;}

while (c != '-' && (c < '' || c > '')){c = getchar();}sgn = (c == '-') ? - : ;ret = (c == '-') ?  : (c - '');

while (c = getchar(), c >= '' && c <= ''){ret = ret *  + (c - '');}ret *= sgn;return ;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;

while(in!='-'&&in!='.'&&(in<''||in>''))in=getchar();if(in=='-'){IsN=true;num=;}else if(in=='.'){IsD=true;num=;}

else num=in-'';if(!IsD){while(in=getchar(),in>=''&&in<=''){num*=;num+=in-'';}}

if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>=''&&in<=''){num+=Dec*(in-'');Dec*=0.1;}}

if(IsN) num=-num;return true;}

void Out(LL a){if(a < ) { putchar('-'); a = -a; }if(a >= ) Out(a / );putchar(a %  + '');}

void print(LL a){ Out(a),puts("");}

//freopen( "in.txt" , "r" , stdin );

//freopen( "data.txt" , "w" , stdout );

//cerr << "run time is " << clock() << endl;

/*struct Point

{

    double x, y;

    Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数

    Point(double x = 0, double y = 0) : x(x), y(y) { }

    inline void input()

    {

        scanf("%lf%lf",&x,&y);

    }

    inline void print()

    {

        printf("%.6lf %.6lf\n",x,y);

    }

};*/

db sqr(db x)

{

   return x*x;

}

int dcmp(double x)

{

    if(fabs(x) < EPS) return ;

    else return x <  ? - : ;

}

struct Circle

{

    double x, y, r, angle;

    int d;

    Circle(){}

    Circle(double xx, double yy, double ang = , int t = )

    {

        x = xx;  y = yy;  angle = ang;  d = t;

    }

    void get()

    {

        scanf("%lf%lf%lf", &x, &y, &r);

        d = ;

    }

};

Circle cir[maxn],tp[maxn*];

double area[maxn];

double dis(Circle a,Circle b)

{

    return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));

}

double cross(Circle p0,Circle p1,Circle p2)

{

    return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);

}

//圆相交

int CirCrossCir(Circle p1, double r1,Circle p2, double r2,Circle &cp1,Circle &cp2)

{

    double mx = p2.x - p1.x, sx = p2.x + p1.x, mx2 = mx * mx;

    double my = p2.y - p1.y, sy = p2.y + p1.y, my2 = my * my;

    double sq = mx2 + my2, d = -(sq - sqr(r1 - r2)) * (sq - sqr(r1 + r2));

    if (d + eps < ) return ; if (d < eps) d = ; else d = sqrt(d);

    double x = mx * ((r1 + r2) * (r1 - r2) + mx * sx) + sx * my2;

    double y = my * ((r1 + r2) * (r1 - r2) + my * sy) + sy * mx2;

    double dx = mx * d, dy = my * d; sq *= ;

    cp1.x = (x - dy) / sq; cp1.y = (y + dx) / sq;

    cp2.x = (x + dy) / sq; cp2.y = (y - dx) / sq;

    if (d > eps) return ; else return ;

}

bool circmp(const Circle& u, const Circle& v)

{

    return dcmp(u.r - v.r) < ;

}

bool cmp(const Circle& u, const Circle& v)

{

    if (dcmp(u.angle - v.angle)) return u.angle < v.angle;

    return u.d > v.d;

}

//0.5*r*r*(K-sin(K))

double calc(Circle cir,Circle cp1,Circle cp2)

{

    double ans = (cp2.angle - cp1.angle) * sqr(cir.r)

        - cross(cir, cp1, cp2) + cross(Circle(, ), cp1, cp2);

    return ans / ;

}

void CirUnion(Circle cir[], int n)

{

    Circle cp1, cp2;

    sort(cir, cir + n, circmp);

    for (int i = ; i < n; ++i)

        for (int j = i + ; j < n; ++j)

            if (dcmp(dis(cir[i], cir[j]) + cir[i].r - cir[j].r) <= )

                cir[i].d++;

    for (int i = ; i < n; ++i)

    {

        int tn = , cnt = ;

        for (int j = ; j < n; ++j)

        {

            if (i == j) continue;

            if (CirCrossCir(cir[i], cir[i].r, cir[j], cir[j].r,

                cp2, cp1) < ) continue;

            cp1.angle = atan2(cp1.y - cir[i].y, cp1.x - cir[i].x);

            cp2.angle = atan2(cp2.y - cir[i].y, cp2.x - cir[i].x);

            cp1.d = ;    tp[tn++] = cp1;

            cp2.d = -;   tp[tn++] = cp2;

            if (dcmp(cp1.angle - cp2.angle) > ) cnt++;

        }

        tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, pi, -cnt);

        tp[tn++] = Circle(cir[i].x - cir[i].r, cir[i].y, -pi, cnt);

        sort(tp, tp + tn, cmp);

        int p, s = cir[i].d + tp[].d;

        for (int j = ; j < tn; ++j)

        {

            p = s;  s += tp[j].d;

            area[p] += calc(cir[i], tp[j - ], tp[j]);

        }

    }

}

int n;

void solve()

{

    for(int i=;i<n;i++)

        cir[i].get();

    me(area,);

    CirUnion(cir,n);

    for(int i=;i<=n;i++)

    {

        area[i]-=area[i+];

        printf("[%d] = %.3f\n", i, area[i]);

    }

}

int main()

{

    while(scanf("%d",&n)!=EOF)

        solve();

}
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圆的k次面积并

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