poj 2932 Coneology （扫描线）

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题意

平面上有N个两两不相交的圆，求所有最外层的，即不被其他圆包含的圆的个数并输出

思路

挑战程序竞赛P259页

代码

/* **********************************************Auther: xueaohuiCreated Time: 2015-7-25 16:56:13File 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 111111int 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\n",res.size());    for(int i=0;i<res.size();i++){       if(i!=0) printf(" ");       printf("%d",res[i]+1);    }    printf("\n");}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();    }}

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