Archery Tournament

题意：有n次操作，每次输入三个数x，y，z，若x为1，表示以（y，z）为圆心画一个与x轴相切的圆（保证画出来的圆一定不相交），若x为0，则问（y，z）这个点在哪个圆上，若不在任何圆上，输出-1；若在某个圆上，输出是第几次操作画的圆，并将圆删去

解题思路：线段树+set，因为圆都是不相交的，并且圆心都是整数，所以圆不会很多

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>#include <ctime>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n;int x[200009], y[200009], z[200009];int a[400009], m, tot;set<int>s[400009 << 2];void build(int k, int l, int r){s[k].clear();if (l == r) return;int mid = (l + r) >> 1;build(k << 1, l, mid), build(k << 1 | 1, mid + 1, r);}void update(int k, int l, int r, int ll, int rr, int p, int flag){if (l >= ll&r <= rr){if (flag) s[k].insert(p);else s[k].erase(p);return;}int mid = (l + r) >> 1;if (mid >= ll) update(k << 1, l, mid, ll, rr, p, flag);if (mid < rr) update(k << 1 | 1, mid + 1, r, ll, rr, p, flag);}int query(int k, int l, int r, int p, int id){if (l <= p&&r >= p){for (set<int>::iterator it = s[k].begin(); it != s[k].end(); it++){int temp = *it;if (1LL * z[temp] * z[temp] > 1LL * (z[id] - z[temp])*(z[id] - z[temp]) + 1LL * (y[id] - y[temp])*(y[id] - y[temp]))return temp;}}if (l == r) return -1;int mid = (l + r) >> 1;if (p <= mid) return query(k << 1, l, mid, p, id);else return query(k << 1 | 1, mid + 1, r, p, id);}int main(){while (~scanf("%d", &n)){tot = 1;for (int i = 1; i <= n; i++){scanf("%d %d %d", &x[i], &y[i], &z[i]);if (x[i] == 1) a[tot++] = y[i] - z[i], a[tot++] = y[i] + z[i];else a[tot++] = y[i];}sort(a + 1, a + tot);m = unique(a + 1, a + tot) - a - 1;build(1, 1, m);for (int i = 1; i <= n; i++){if (x[i] == 1){int xx = lower_bound(a + 1, a + m + 1, y[i] - z[i]) - a;int yy = lower_bound(a + 1, a + m + 1, y[i] + z[i]) - a;update(1, 1, m, xx, yy, i, 1);}else{int xx = lower_bound(a + 1, a + m + 1, y[i]) - a;int ans = query(1, 1, m, xx, i);if (ans != -1){int xx = lower_bound(a + 1, a + m + 1, y[ans] - z[ans]) - a;int yy = lower_bound(a + 1, a + m + 1, y[ans] + z[ans]) - a;update(1, 1, m, xx, yy, ans, 0);}printf("%d\n", ans);}}}return 0;}