HDU 3511 圆扫描线

找最深的圆，输出层数

类似POJ 2932的做法 圆扫描线即可。这里要记录各个圆的层数，所以多加一个维护编号的就行了。

/** @Date    : 2017-10-18 18:16:52
  * @FileName: HDU 3511 圆扫描线.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.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;

struct yuu
{
	double x, y, r;
	yuu(){}
	yuu(double _x, double _y, double _r):x(_x), y(_y), r(_r){}
};
yuu a[N];
double scanx;

struct cir
{
	int m;
	int flag;
	cir(){}
	cir(int _m, int _f):m(_m), flag(_f){}
	bool operator <(const cir &b) const{
		double dis1 = sqrt(a[m].r * a[m].r - (scanx - a[m].x) * (scanx - a[m].x));
		double dis2 = sqrt(a[b.m].r * a[b.m].r - (scanx - a[b.m].x) * (scanx - a[b.m].x));
		double y1 = a[m].y + dis1 * flag;
		double y2 = a[b.m].y + dis2 * b.flag;
		return y1 < y2|| (fabs(y1 - y2) < eps && flag < b.flag);
	}
};

pair<double ,int> p[N*2];

int ans[N];

int main()
{
	int n;
	while(cin >> n)
	{
		MMF(ans);
		for(int i = 0; i < n; i++)
		{
			double x, y, r;
			scanf("%lf%lf%lf", &x, &y, &r);
			a[i] = yuu(x, y, r);
			p[i*2] = MP(x - r, i);
			p[i*2 + 1] = MP(x + r, i + n);
		}
		sort(p, p + 2 * n);
		set<cir>st;//按交点高度维护圆集合
		int ma = 0;
		for(int i = 0; i < 2 * n; i++)
		{
			scanx = p[i].fi;
			if(p[i].se < n)
			{
				int up = -1, dw = -1;
				st.insert(cir(p[i].se, -1));
				auto pos = st.lower_bound(cir(p[i].se, -1));
				if((++pos) != st.end())
					up = (pos)->m;
				if((--pos) != st.begin())
					dw = (--pos)->m;
				if(up == dw && ~up)
					ans[p[i].se] = ans[up] + 1;
				else if(~up && ~dw)
					ans[p[i].se] = max(ans[up], ans[dw]);
				else ans[p[i].se] = 1;
				//cout << up << dw << endl;
				st.insert(cir(p[i].se, 1));
				ma = max(ma, ans[p[i].se]);
			}
			else
			{
				st.erase(cir(p[i].se%n, 1));
				st.erase(cir(p[i].se%n, -1));
			}
		}
		printf("%d\n", ma);
	}
    return 0;
}