Codeforces Beta Round #12 (Div 2 only) D. Ball tree array query suffix, max

Http://codeforces.com/problemset/problem/12/D

The bit query here refers to the maximum value between the query [1, R] or [R, Maxn], which is sufficient.

Set three weights respectively b[1], b[2], b[2];

First, the b[1] value discretization, discrete into a ID, then can only be found in the location of the larger ID. Then the b[2] sort, reverse query, that is, the first query the largest, of course, it is not possible to commit suicide, because it has the largest, and then after the query, put it into a bit, after the bit query, do not need to tube b[2], because in bit inside the b[2] is always the larger, In fact, this is a common method. Hack point is the same value can not be updated in advance, the details see Code + slowly debug

Now talk about the role of bit, bit maintenance is b[3], because we have discretized b[1], then each element of the subscript is equivalent to the discretization of the b[1], so add in the bit when the ID do subscript. The bit maintains the suffix maximum value.

Hack point means, if your a[i].b[2] = = a[j].b[2], then can not be in the query A[j] ahead of the a[i] to join bit, because bit query by default is B[2] established, but in advance to update the words, not set.

#include <cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<assert.h>#defineIOS Ios::sync_with_stdio (False)using namespacestd;#defineINF (0X3F3F3F3F)typedefLong Long intLL; #include<iostream>#include<sstream>#include<vector>#include<Set>#include<map>#include<queue>#include<string>#include<bitset>#include<stack>Const intMAXN = 5e5 + -;intC[MAXN], en;intLowbit (intx) {returnX & (-x);}voidUpDate (intPosintval) {     while(POS) {C[pos]=Max (C[pos], Val); POS-=Lowbit (POS); }}intAskintPOS) {    intAns =-1;  while(Pos <=en) {ans=Max (ans, c[pos]); POS+=Lowbit (POS); }    returnans;}structNode {inta[5], id;} A[maxn];stack<int>St;BOOLCmp0 (structNode A,structNode B) {    returna.a[1] < b.a[1];}BOOLCMP1 (structNode A,structNode B) {    returna.a[2] < b.a[2];}voidWork () {intN; scanf ("%d", &N);  for(inti =1; I <= N; ++i) scanf ("%d", &a[i].a[1]);  for(inti =1; I <= N; ++i) scanf ("%d", &a[i].a[2]);  for(inti =1; I <= N; ++i) scanf ("%d", &a[i].a[3]); Sort (a+1, A +1+N, cmp0); En=1; a[1].id =en;  for(inti =2; I <= N; ++i) {if(a[i].a[1] = = A[i-1].a[1]) {a[i].id=en; } ElseA[i].id = + +en; } en+=2; Memset (c,-1,sizeofc); Sort (a+1, A +1+N, CMP1);//for (int i = 1; I <= n; ++i) {//cout << a[i].a[1] << "<< a[i].a[2] <<" "<< a[i].a[3] <<" "<< A[i].id << Endl;//    }St.push (n); intAns =0;  for(inti = n-1; I >=1; --i) {if(a[i].a[2] = = A[i +1].a[2]) {St.push (i); intres = Ask (A[i].id +1); Ans+ = res > a[i].a[3]; } Else {             while(!St.empty ()) {                intres =St.top ();                St.pop (); UpDate (a[res].id, a[res].a[3]); }            intres = Ask (A[i].id +1); Ans+ = res > a[i].a[3]; St.push (i);//upDate (A[i].id, a[i].a[3]);}} cout<< ans <<Endl;}intMain () {#ifdef local freopen ("Data.txt","R", stdin);//freopen ("Data.txt", "w", stdout);#endifWork (); return 0;}

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Codeforces Beta Round #12 (Div 2 only) D. Ball tree array query suffix, max