poj 2932 Coneology

最新推荐文章于 2020-10-02 19:30:05 发布

quinwu-

最新推荐文章于 2020-10-02 19:30:05 发布

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本文链接：https://blog.youkuaiyun.com/acm_wukuan/article/details/41578063

圆的一道扫描线的题

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<string>
#include<set>
#include<vector>
using namespace std;
const int maxn = 40010;

int N;
double x[maxn],y[maxn],r[maxn];

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

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

    set<pair<double,int> > outers;
    vector<int> res;

    for(int i  = 0 ; i < events.size() ; i++)
    {
        int id = events[i].second % N;
        if(events[i].second < N)
        {
            set<pair<double,int> >::iterator it = outers.lower_bound(make_pair(y[id],id));

            if(it != outers.end() && inside(id,it->second)) continue;
            if(it != outers.begin() && inside(id,(--it)->second)) continue;
            res.push_back(id);
            outers.insert(make_pair(y[id],id));
        }
        else
        {
            outers.erase(make_pair(y[id],id));
        }
    }

    sort(res.begin(),res.end());
    printf("%d\n",res.size());
    for(int i = 0 ; i < res.size(); i++)
    {
        printf("%d ",res[i] +1);
    }
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    scanf("%d",&N);
    for(int i = 0 ; i < N ; i++)
    {
        scanf("%lf %lf %lf",&r[i],&x[i],&y[i]);
    }
    solve();
    return 0;
}