[hdu4631 Sad Love Story]最近点对，枚举

题意：S是平面内点的集合，初始为空，每次向集合里面加入一个点P(x,y)，询问S内最近点对的距离的平方和

思路：设当前集合的答案为D，则找到集合里面横坐标在(x-√D,x+√D)内的数，用它们来更新答案，一边更新答案一边还要更新右边界x+√D，此时的更新注意不要用浮点数开平方算具体右边界，改用判断即可，否则超时。

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#include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} template<typename T> void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} template<typename T> void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ multiset<pii> S; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, n, ax, bx, cx, ay, by, cy, x, y; cin >> T; while (T --) { scanf("%d%d%d%d%d%d%d", &n, &ax, &bx, &cx, &ay, &by, &cy); S.clear(); x = bx % cx; y = by % cy; S.insert(mp(x, y)); ll ans = 0, mind = (ll)INF * INF; for (int i = 1; i < n; i ++) { x = ((ll)x * ax + bx) % cx; y = ((ll)y * ay + by) % cy; int dx = (int)(sqrt(mind * 1.0) + 1); multiset<pii>::iterator iter = S.lower_bound(mp(x - dx, -INF)); for (; iter != S.end(); iter ++) { ll X = (*iter).X - x, Y = (*iter).Y - y; ll buf = X * X + Y * Y; if (X >= 0 && X * X >= mind) break; umin(mind, buf); } ans += mind; S.insert(mp(x, y)); } cout << ans << endl; } return 0; }