SPOJ - CIRU - The area of the union of circles / BZOJ 2178

CIRU - The area of the union of circles

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You are given N circles and expected to calculate the area of the union of the circles !

Input

The first line is one integer n indicates the number of the circles. (1 <= n <= 1000)

Then follows n lines every line has three integers

Xi Yi Ri

indicates the coordinate of the center of the circle, and the radius. (|Xi|. |Yi|  <= 1000, Ri <= 1000)

Note that in this problem Ri may be 0 and it just means one point !

Output

The total area that these N circles with 3 digits after decimal point

Example

Input:
3

0 0 1
0 0 1
100 100 1


Output:
6.283
题意：求圆的面积并
解法：simpson积分，递归至左右区间积分和与大区间积分的差值小于EPS

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>
 
using namespace std;
 
#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define pli pair<ll, int>
#define pil pair<int, ll>
#define pdd pair<double, double>
#define clr(a, x) memset(a, x, sizeof(a))
 
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
 
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define TREE tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update>
TREE T;
*/
 
struct Circle {
    double x, y, r;
    Circle () {}
    Circle (double x, double y, double r) : x(x), y(y), r(r) {}
} c[1005];
int n;
double Cut(double x) {
    pdd arr[1005];
    int index = 0;
    for (int i = 1; i <= n; ++i) {
        if (fabs(x - c[i].x) < c[i].r) {
            double dd = sqrt(c[i].r * c[i].r - (x - c[i].x) * (x - c[i].x));
            arr[++index] = pdd(c[i].y - dd, c[i].y + dd);
        }
    }
    sort(arr + 1, arr + index + 1);
    double res = 0, ma = -INF;
    for (int i = 1; i <= index; ++i) {
        res += max(0.0, arr[i].second - max(arr[i].first, ma));
        ma = max(ma, arr[i].second);
    }
    return res;
}
double simpson(double l, double r, double fl, double fr) {
    double fmi = Cut((l + r) * 0.5);
    return (r - l) / 6.0 * (fl + 4 * fmi + fr);
}
double solve(double l, double r) {
    double mi = (l + r) * 0.5;
    double fmi = Cut(mi), fl = Cut(l), fr = Cut(r);
    double ans = simpson(l, r, fl, fr);
    double ans1 = simpson(l, mi, fl, fmi);
    double ans2 = simpson(mi, r, fmi, fr);
    if (r - l < 1 && fabs(ans1 + ans2 - ans) < EPS) return ans1 + ans2;
    return solve(l, mi) + solve(mi, r);
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lf%lf%lf", &c[i].x, &c[i].y, &c[i].r);
    }
    printf("%.3f\n", solve(-2000, 2000));    
    return 0;
}