http://poj.org/problem?id=2932
先給圓的最左端和最右端的點排個序,當兩點x相同時,左端點排在前面。
然後就是掃描了,
若掃描到的是圓的左端點,就判斷圓心(y座標)在其上方且離其最近的圓是否包含此圓,以及圓心(y座標)在其下方且離其最近的圓是否包含此圓,若包含就continue,不包含就insert到set中;
若掃描到的是圓的右端點,就從set中erase此圓(erase操作考慮了圓不在set中的情況)
完整代碼:
/*2282ms,5064KB*/
#include<cstdio>
#include<set>
#include<vector>
#include<utility>
#include<algorithm>
using namespace std;
const int mx = 40005;
int n;
double x[mx], y[mx], r[mx];
pair<double, int> px[mx * 2];
vector<int> res; ///答案
set<pair<double, int> > out; ///最外層的圓的集合(維護圓心縱座標)
set<pair<double, int> >::iterator it;
///判斷圓i是否在圓j內部
inline bool inside(int i, int j)
{
return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) <= r[j] * r[j];
}
void solve()
{
for (int i = 0, j = 0; i < n; ++i)
{
px[j++] = make_pair(x[i] - r[i], i);
px[j++] = make_pair(x[i] + r[i], i + n);
}
int m = n * 2;
sort(px, px + m);
for (int i = 0; i < m; ++i)
{
int id = px[i].second % n;
if (px[i].second < n) ///掃描到左端
{
it = out.lower_bound(make_pair(y[id], id));
if (it != out.end() && inside(id, it->second) ||
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)); ///掃描到右端
}
sort(res.begin(), res.end());
printf("%d\n", res.size());
printf("%d", res[0] + 1);
for (int i = 1; i < res.size(); ++i) printf(" %d", res[i] + 1);
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%lf%lf%lf", &r[i], &x[i], &y[i]);
solve();
return 0;
}