求n个圆中没有被包含的圆。模仿扫描线从左往右扫，到左边界此时如有3个交点，则有3种情况，以此判定该圆是否被离它最近的圆包含，而交点和最近的圆可以用以y高度排序的Set来维护。因此每次到左边界插入该圆，找该圆最近的两个圆（上方和下方）判断是否包含，到右边界则从Set中删除该圆。

/** @Date    : 2017-08-13 17:27:55* @FileName: POJ 2932 圆扫描线.cpp* @Platform: Windows* @Author  : Lweleth (SoungEarlf@gmail.com)* @Link    : https://github.com/* @Version : $Id$*/
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;int n;
struct circle
{double x, y, r;circle(){}circle(double _x, double _y, double _r){x = _x, y = _y, r = _r;}
};circle c[N];
pair<double, int>p[N*2];
int ans[N];int isinside(int a, int b)
{double dx = c[a].x - c[b].x;double dy = c[a].y - c[b].y;return dx * dx + dy * dy <= c[b].r * c[b].r;
}int main()
{scanf("%d", &n);for(int i = 0; i < n; i++){double x, y, r;scanf("%lf%lf%lf", &r, &x, &y);c[i] = circle(x, y, r);}for(int i = 0; i < n; i++){p[i*2] = MP(c[i].x - c[i].r, i);p[i*2 + 1] = MP(c[i].x + c[i].r, i + n);}sort(p, p + 2 * n);set<pair<double, int> >q;int cnt = 0;for(int i = 0; i < 2 * n; i++){if(p[i].se < n){set<pair<double, int> >::iterator it;/*auto */it = q.lower_bound(MP(c[p[i].se].y, p[i].se));if(it != q.end() && isinside(p[i].se, it->se)|| it != q.begin()&&isinside(p[i].se, (--it)->se))continue;q.insert(MP(c[p[i].se].y, p[i].se));ans[cnt++] = p[i].se;}else q.erase(MP(c[p[i].se%n].y, p[i].se%n)); }sort(ans, ans + cnt);printf("%d\n", cnt);for(int i = 0; i < cnt; i++)printf("%d%s", ans[i] + 1, i==cnt-1?"\n":" ");return 0;
}