给出n个点，计算从0开始每加入一个点最近点对点的距离的平方，累加答案
思路：用模板直接暴力复杂度太高，考虑平衡树来支持插入操作（set）
由于此题只涉及点的操作，无需写大量计算几何模板，直接利用分治求最近点对，加入以后check最近的即可
#include<iostream> #include<unordered_map> #include<algorithm> #include<string> #include<cmath> #include<cstring> #include<vector> #include<map> #include<set> #include<queue> #include<deque> #include<stack> #include<sstream> #include <cstdlib> #include<cstdio> #include<random> #define INF 0x3f3f3f3f #define eps 1e-8 #define equals(a,b) (fabs(a-b)<eps) #define MOD 100000007 std::mt19937 rnd(233); //#define rnd() rand()%1000 const int maxn = 5e5+5 ; const double PI = acos(-1.0); typedef long long ll; using namespace std; struct Point { ll x; ll y; Point(ll x = 0, ll y = 0) :x(x), y(y) {} friend bool operator < (const Point& a, const Point& b) { if (a.x != b.x) return a.x < b.x; else return a.y < b.y; } }; set<Point> v; int main() { int T; int n, a1, b1, c1, a2, b2, c2; scanf("%d", &T); while (T--) { v.clear(); scanf("%d%d%d%d%d%d%d", &n, &a1, &b1, &c1, &a2, &b2, &c2); ll x, y, sum, mids; x = y = sum = 0; mids = (1ll<<60)-1; for (int i = 0; i < n; i++) { x = (x * a1 + b1) % c1; y = (y * a2 + b2) % c2; Point p; p.x = x, p.y = y; if (!v.count(p)) { if (i > 0) { auto it = v.lower_bound(p); for (auto e = it; e != v.end(); e++) { int next_x = e->x - p.x; if (next_x * next_x > mids) break; int next_y = e->y - p.y; ll dis = (ll)next_x * next_x + (ll)next_y * next_y; mids = min(mids, dis); } for (auto e = it; e != v.begin(); ) { e--; int next_x = e->x - p.x; if (next_x * next_x > mids) break; int next_y = e->y - p.y; ll dis = (ll)next_x * next_x + (ll)next_y * next_y; mids = min(dis, mids); } sum += mids; } v.insert(Point(x, y)); } else break; } printf("%lld\n", sum); } return 0; }