Codeforces Round #418 (Div. 2) D. An overnight dance in discotheque

http://codeforces.com/contest/814/problem/D

题意：给你n个圆的坐标和半径

          你将n个圆分为两部分

         每部分中的圆若被覆盖了奇数次需要减去 偶数次需要加上

思路：贪心

        先把没有被覆盖圆放入上半场

        再把剩下的圆放入下半场

        计算两者的面积和

 ACcode：

#include<bits/stdc++.h>using namespace std;const int maxn = 1100;const double eps = 1e-9;const double pi = acos(-1.0);struct node{    double x, y, r;    node(double xx = 0.0, double yy = 0.0, double rr = 0.0) : x(xx), y(yy), r(rr) {}} del[maxn];int cheak(int a, int b){    double ret1 = (del[a].x - del[b].x) * (del[a].x - del[b].x) + (del[a].y - del[b].y) * (del[a].y - del[b].y);    double ret2 = (del[a].r + del[b].r) * (del[a].r + del[b].r);    if((ret1 - ret2) < 0 && (del[a].r - del[b].r) > 0)        return 1;    else        return 0;}int iv[maxn];int main(){    int n;    scanf("%d", &n);    for(int i = 1; i <= n; i++)        scanf("%lf%lf%lf", &del[i].x, &del[i].y, &del[i].r);    for(int i = 1; i <= n; i++)    {        for(int j = 1; j <= n; j++)        {            if(i == j) continue;            if(cheak(i,j) == 1)                iv[j]++;        }    }    double ans = 0.0;    for(int i=1;i<=n;i++)    {        //printf("i %d  del %f %f %f  iv %d\n",i,del[i].x, del[i].y, del[i].r,iv[i]);        if(iv[i] == 0  || iv[i] % 2 == 1)            ans+=del[i].r*del[i].r*pi;///删除没有被覆盖的圆后 下半场的奇偶互换        else            ans-=del[i].r*del[i].r*pi;        ans+=eps;        ans-=eps;    }    printf("%.9f\n",ans);    return 0;}