題目大意

　　有$n$個元素，第$i$個元素有$a_i,b_i,c_i$三個屬性，設(shè)$f(i)$為$a_j<a_i,b_j<b_i,c_j<c_i$同時滿足的數(shù)量。對于$d\in [0,n)$，求$f(i)=d$的數(shù)量。

思路

#include <cstdio> #include <cstring> #include <algorithm> using namespace std;const int MAX_ELEMENT = 100010, MAX_MAXVAL = 200010; int TotEle, MaxVal; int F[MAX_MAXVAL];struct Element {int A, B, C;bool operator == (const Element& a) const{return A == a.A && B == a.B && C == a.C;}bool operator < (const Element& a) const{return A != a.A ? A < a.A : B != a.B ? B < a.B : C < a.C;} }_eles[MAX_ELEMENT];//elementsstruct RangeTree { private:struct Node{Node *LeftSon, *RightSon;int Cnt;Node():LeftSon(NULL), RightSon(NULL), Cnt(0){}}*Root;void Update(Node *&cur, int l, int r, int p, int delta){if (!cur)cur = new Node();cur->Cnt += delta;if (l == r)return;int mid = (l + r) / 2;if (p <= mid)Update(cur->LeftSon, l, mid, p, delta);if (p > mid)Update(cur->RightSon, mid + 1, r, p, delta);}int Query(Node *cur, int sl, int sr, int al, int ar){if (!cur)return 0;if (al <= sl && sr <= ar)return cur->Cnt;int ans = 0, mid = (sl + sr) / 2;if (al <= mid)ans += Query(cur->LeftSon, sl, mid, al, ar);if (ar > mid)ans += Query(cur->RightSon, mid + 1, sr, al, ar);return ans;}public:RangeTree():Root(NULL){}void operator += (int p){Update(Root, 1, MaxVal, p, 1);}int Query(int l, int r){return Query(Root, 1, MaxVal, l, r);} };struct BIT { private:RangeTree C[MAX_MAXVAL];int N;int Lowbit(int x){return x & -x;}public:void Init(int n){N = n;}void Update(int p, int delta){while (p <= N){C[p] += delta;p += Lowbit(p);}}int Query(int p, int qKey){int ans = 0;while (p > 0){ans += C[p].Query(1, qKey);p -= Lowbit(p);}return ans;} }g;void Solve() {sort(_eles + 1, _eles + TotEle + 1);int prevCnt = 1;for (int i = 1; i <= TotEle; i++){if (_eles[i] == _eles[i + 1]){prevCnt++;g.Update(_eles[i].B, _eles[i].C);continue;}F[g.Query(_eles[i].B, _eles[i].C)] += prevCnt;g.Update(_eles[i].B, _eles[i].C);prevCnt = 1;} }int main() {scanf("%d%d", &TotEle, &MaxVal);g.Init(MaxVal);for (int i = 1; i <= TotEle; i++)scanf("%d%d%d", &_eles[i].A, &_eles[i].B, &_eles[i].C);Solve();for (int i = 0; i <= TotEle - 1; i++)printf("%d\n", F[i]);return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/headboy2002/p/9375413.html

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