poj 2932 Coneology （扫描线）

题意

平面上有N个两两不相交的圆，求全部最外层的，即不被其它圆包括的圆的个数并输出

思路

挑战程序竞赛P259页

代码

/* **********************************************
Auther: xueaohui
Created Time: 2015-7-25 16:56:13
File Name   : poj2932.cpp
*********************************************** */
#include <iostream>
#include <fstream>
#include <cstring>
#include <climits>
#include <deque>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <functional>
#include <algorithm>
using namespace std;
#define ll long long
#define N 111111
int n;
double x[N],y[N],r[N];

bool inside(int i,int j){
    double px = x[i]-x[j];
    double py = y[i]-y[j];
    return px*px+py*py<=r[j]*r[j];
}

void slove(){
    vector<pair<double,int>>e;
    e.clear();
    for(int i=0;i<n;i++){
        e.push_back(make_pair(x[i]-r[i],i));
        e.push_back(make_pair(x[i]+r[i],i+n));
    }
    sort(e.begin(),e.end());

    set<pair<double,int>>out;

    vector<int>res;
    res.clear();
    out.clear();

    for(int i=0;i<e.size();i++){
        int id = e[i].second %n;
        if(e[i].second<n){
            set<pair<double,int>>::iterator it = out.lower_bound(make_pair(y[id],id));
            if(it != out.end() && inside (id,it->second)) continue;
            if(it != out.begin() && inside (id ,(--it)->second)) continue;
            res.push_back(id);
            out.insert(make_pair(y[id],id));
        }
        else{
            out.erase(make_pair(y[id],id));
        }
    }
    sort(res.begin(),res.end());
    printf("%d
",res.size());
    for(int i=0;i<res.size();i++){
       if(i!=0) printf(" ");
       printf("%d",res[i]+1);
    }
    printf("
");
}

int main(){
    while(scanf("%d",&n)==1){
    for(int i=0;i<n;i++){
        scanf("%lf%lf%lf",&r[i],&x[i],&y[i]);
    }
    slove();
    }
}