[SPOJ][8119][CIRU2][圆的面积并]

/*
题目：CIRU2
题目来源：SPOJ 8119  https://www.spoj.pl/problems/CIRUT/
题目难度：中等偏难
题目内容或思路：
    圆的面积并
    题意：给n个圆（最多1000个），分别求出覆盖1层的面积、覆盖2层的面积、
        覆盖3层的面积。。。覆盖n层的面积
    方法见AC大牛blog：
    http://hi.baidu.com/aekdycoin/blog/item/c1b28e3711246b3f0b55a95e.html
做题日期：2011.3.27
*/
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <vector>
#include <bitset>
#include <cmath>
#include <set>
#include <utility>
#include <ctime>
#define sqr(x) ((x)*(x))
using namespace std;

const int N = 1010;
const double eps = 1e-8;
const double pi = acos(-1.0);
double area[N];
int n;

int dcmp(double x) {
    if (x < -eps) return -1; else return x > eps;
}

struct cp {
    double x, y, r, angle;
    int d;
    cp(){}
    cp(double xx, double yy, double ang = 0, int t = 0) {
        x = xx;  y = yy;  angle = ang;  d = t;
    }
    void get() {
        scanf("%lf%lf%lf", &x, &y, &r);
        d = 1;
    }
}cir[N], tp[N * 2];

double dis(cp a, cp b) {
    return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}

double cross(cp p0, cp p1, cp p2) {
    return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);
}

int CirCrossCir(cp p1, double r1, cp p2, double r2, cp &cp1, cp &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 < 0) return 0; if (d < eps) d = 0; 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 *= 2;
    cp1.x = (x - dy) / sq; cp1.y = (y + dx) / sq;
    cp2.x = (x + dy) / sq; cp2.y = (y - dx) / sq;
    if (d > eps) return 2; else return 1;
}

bool circmp(const cp& u, const cp& v) {
    return dcmp(u.r - v.r) < 0;
}

bool cmp(const cp& u, const cp& v) {
    if (dcmp(u.angle - v.angle)) return u.angle < v.angle;
    return u.d > v.d;
}

double calc(cp cir, cp cp1, cp cp2) {
    double ans = (cp2.angle - cp1.angle) * sqr(cir.r) 
        - cross(cir, cp1, cp2) + cross(cp(0, 0), cp1, cp2);
    return ans / 2;
}

void CirUnion(cp cir[], int n) {
    cp cp1, cp2;
    sort(cir, cir + n, circmp);
    for (int i = 0; i < n; ++i)
        for (int j = i + 1; j < n; ++j)
            if (dcmp(dis(cir[i], cir[j]) + cir[i].r - cir[j].r) <= 0)
                cir[i].d++;
    for (int i = 0; i < n; ++i) {
        int tn = 0, cnt = 0;
        for (int j = 0; j < n; ++j) {
            if (i == j) continue;
            if (CirCrossCir(cir[i], cir[i].r, cir[j], cir[j].r,
                cp2, cp1) < 2) 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 = 1;    tp[tn++] = cp1;
            cp2.d = -1;   tp[tn++] = cp2;
            if (dcmp(cp1.angle - cp2.angle) > 0) cnt++;
        }
        tp[tn++] = cp(cir[i].x - cir[i].r, cir[i].y, pi, -cnt);
        tp[tn++] = cp(cir[i].x - cir[i].r, cir[i].y, -pi, cnt);
        sort(tp, tp + tn, cmp);
        int p, s = cir[i].d + tp[0].d;
        for (int j = 1; j < tn; ++j) {
            p = s;  s += tp[j].d;
            area[p] += calc(cir[i], tp[j - 1], tp[j]);
        }
    }
}

void solve() {
    for (int i = 0; i < n; ++i)
        cir[i].get();
    memset(area, 0, sizeof(area));
    CirUnion(cir, n);
    for (int i = 1; i <= n; ++i) {
        area[i] -= area[i + 1];
        printf("[%d] = %.3lf\n", i, area[i]);
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("D:\\in.txt", "r", stdin);
#endif
    while (scanf("%d", &n) != EOF) {
        solve();
    }
    return 0;
}