HDU 5773 the all-purpose Zero longest ascending subsequence + tree array

Title Link: Hdu 5773 the all-purpose Zero

The official puzzle: 0 can be converted to any integer, including negative numbers, obviously to find the LIS as far as possible to put 0 in must be correct.

So we can take 0 out of the remaining to do O (Nlogn) of the LIS, the statistical results of the time to calculate the number of 0.

In order to guarantee the strict increment, we can subtract each weight value s[i] minus I front 0 number, then do LIS, can guarantee the result is strictly increment.

Personal view: For the obvious put so 0 in the part I explained:

• If the 0 bits are on either side of the longest ascending subsequence, the two 0 must be added in the obvious
• If there is a 0 sandwiched between the longest ascending sub-sequence, then the number of the 0 left in the LIS should be smaller than the number of 0 to the right 1, and because of the preprocessing effect, the number of the right is at least 2 greater than the number on the left, so 0 can be inserted, for example: 1 2 3 0 4 5 6-> after preprocessing: 1 2 3 0 3 4 5, the longest is 1 2 3 0 4 (5) 5 (6), in parentheses is the original number, in this case 0 is the replacement of the original sequence 0 to the right of 4
• There are a lot of 0 clips in Lis, the same can be

/************************************************************** problem:hdu 5773 the All-purpose Zero User:youmi language:c++ result:accepted time:171ms memory:3148k******************************************************** ********///#pragma COMMENT (linker, "/stack:1024000000,1024000000")//#include <bits/stdc++.h>#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<stack>#include<Set>#include<sstream>#include<cmath>#include<queue>#include<deque>#include<string>#include<vector>#defineZeros (a) memset (A,0,sizeof (a))#defineOnes (a) memset (A,-1,sizeof (a))#defineSC (a) scanf ("%d", &a)#defineSC2 (A, b) scanf ("%d%d", &a,&b)#defineSC3 (a,b,c) scanf ("%d%d%d", &a,&b,&c)#defineSCS (a) scanf ("%s", a)#defineSclld (a) scanf ("%i64d", &a)#definePT (a) printf ("%d\n", a)#definePtlld (a) printf ("%i64d\n", a)#defineRep (i,from,to) for (int i=from;i<=to;i++)#defineIrep (I,to,from) for (int i=to;i>=from;i--)#defineMax (a) (a) > (b)? ( A):(B))#defineMin (a) < (b) ( A):(B))#defineLson (step<<1)#defineRson (lson+1)#defineEPS 1e-6#defineOO 0x3fffffff#defineTEST cout<< "*************************" <<endlConst DoublePi=4*atan (1.0);using namespaceStd;typedefLong Longll;template<classT> InlinevoidRead (T &N) {    CharCintFlag =1;  for(c = GetChar ();! (c >='0'&& C <='9'|| c = ='-'); c = GetChar ());if(c = ='-') flag =-1, n =0;Elsen = C-'0';  for(c = GetChar (); C >='0'&& C <='9'; c = GetChar ()) n = n *Ten+ C-'0'; N *=Flag;}intPow (int Base, ll N,intmo) {    if(n = =0)return 1; if(n = =1)return Base%mo; intTMP = Pow (Base, N >>1, MO); TMP= (LL) TMP * TMP%mo; if(N &1) TMP = (LL) TMP *Base%mo; returntmp;}//***************************intN;Const intmaxn=200000+Ten;intA[MAXN];intVAL[MAXN];intC[MAXN];inttot;intLowbit (intx) {    returnx& (-x);}voidUpdateintTempintx) {     while(x<=tot) {C[x]=Max (c[x],temp); X+=lowbit (x); }}intQueryintx) {    intres=0;  while(x) {res=Max (c[x],res); X-=lowbit (x); }    returnRes;}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin); #endif    intt_t; scanf ("%d",&t_t);  for(intKase=1; kase<=t_t;kase++) {SC (n);        Zeros (c); intCnt=0; inttt=0; intu; Rep (I,1, N)            {SC (u); if(u==0) CNT++; Else{A[tt]=u-CNT; VAL[TT]=A[TT]; TT++; }} sort (Val,val+TT); Tot=unique (VAL,VAL+TT)-Val; intans=0; Rep (I,0, tt-1)        {            intTemp=lower_bound (Val,val+tot,a[i])-val+1; intRes=query (temp-1)+1; Ans=Max (ans,res);        Update (RES,TEMP); } printf ("Case #%d:%d\n", kase,ans+CNT); }}

HDU 5773 the all-purpose Zero longest ascending subsequence + tree array