Hdu_3308_ Segment Tree _ interval Merging

Source: Internet
Author: User

LCIs

Time limit:6000/2000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 6166 Accepted Submission (s): 2675


Problem descriptiongiven n integers.
You have both operations:
U A B:replace The Ath number by B. (index counting from 0)
Q A b:output The length of the longest consecutive increasing subsequence (LCIs) in [A, B].

Inputt in the first line, indicating the case number.
Each case is starts with the integers n, m (0<n,m<=105).
The next line has n integers (0<=val<=105).
The next m lines each have an operation:
U A B (0<=a,n, 0<=b=105)
OR
Q A B (0<=a<=b< N).

Outputfor each Q, output the answer.

Sample Input110 107 7 3 3 5 9 9 8 1 8 Q 6 6U 3 4Q 0 1Q 0 5Q 4 7Q 3 5Q 0 2Q 4 6U 6 10Q 0 9

Sample Output11423125 The first interval of the segment tree to merge.

See the puzzle, from the afternoon to do the evening ... However, it is a simple interval merger ... Sang-xin ...

Gives a sequence that can be updated in a single point to find the longest continuous increment of the length of a child column.

Each section of the record

struct Node
{
int L,r,len; Left and right end points in a segment tree, and sequence length
int ln,rn; The sequence on this segment is the left and right end
int lm,rm,nm; Contains the left endpoint, which contains the right endpoint, and the longest increment continuous child column length of the whole segment
} tree[maxn<<2];

The length (nm) of the longest consecutive incrementing child column in a sequence (RT) is Max (TREE[RT<<1].NM,TREE[RT<<1|1].NM) and satisfies tree[rt<<1].rn<tree[rt <<1|1].LN (left child's right endpoint small less right operand child's left end) in the condition of the larger of the TREE[RT<<1].RM+TREE[RT<<1|1].LM. After the update point pushup and query are using this idea.

The first wave timeout is due to mid= (L+R)/2, using bit arithmetic faster, mid= (l+r) >>1

There was a problem with the query function writing before.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;#defineMAXN 100005#defineLson l,mid,rt<<1#defineRson mid+1,r,rt<<1|1structnode{intL,r,len; intLn,rn; intlm,rm,nm;} TREE[MAXN<<2];intnum[100005];voidPushup (intRT) {Tree[rt].ln=tree[rt<<1].ln; Tree[rt].rn=tree[rt<<1|1].rn; TREE[RT].LM=tree[rt<<1].LM; Tree[rt].rm=tree[rt<<1|1].rm; TREE[RT].NM=max (tree[rt<<1].nm,tree[rt<<1|1].nm); if(tree[rt<<1].rn<tree[rt<<1|1].ln) {if(tree[rt].lm==tree[rt<<1].len) Tree[rt].lm+=tree[rt<<1|1].LM; if(tree[rt].rm==tree[rt<<1|1].len) tree[rt].rm+=tree[rt<<1].rm; TREE[RT].NM=max (tree[rt].nm,tree[rt<<1].rm+tree[rt<<1|1].lm); }}voidBuildintLintRintRT) {TREE[RT].L=M; TREE[RT].R=R; Tree[rt].len=r-l+1; if(l==r) {TREE[RT].LM=tree[rt].rm=tree[rt].nm=1; Tree[rt].ln=tree[rt].rn=Num[l]; return; }    intMid= (l+r) >>1;    Build (Lson);    Build (Rson); Pushup (RT);}voidUpdateintPosintXintLintRintRT) {    if(l==pos&&r==POS) {Tree[rt].ln=tree[rt].rn=x; return; }    intMid= (l+r) >>1; if(pos<=mid) Update (Pos,x,lson); ElseUpdate (Pos,x,rson); Pushup (RT);}intQueryintLintRintLintRintRT) {    if(l==l&&r==R)returntree[rt].nm; intMid= (l+r) >>1; if(r<=mid)returnquery (L,r,lson); Else if(l>mid)returnquery (L,r,rson); Else    {        intLl=query (L,mid,lson), ans=0; intRr=query (mid+1, R,rson); if(tree[rt<<1].rn<tree[rt<<1|1].ln) ans=min (mid-l+1,tree[rt<<1].RM) +min (r-mid,tree[rt<<1|1].lm); returnMax (Ans,max (LL,RR)); }}/*int query (int l,int r,int i)//query the largest lcis{if (tree[i].l>=l && tree[i].r<=r) {return tree[i].    Nm    } int mid = (TREE[I].L+TREE[I].R) >>1,ans = 0;    if (l<=mid) ans = max (ans,query (l,r,2*i));    if (r>mid) ans = max (ans,query (l,r,2*i+1));    if (Tree[2*i].rn < tree[2*i+1].ln) ans = max (ans, min (mid-l+1,tree[2*i].rm) +min (R-MID,TREE[2*I+1].LM)); return ans;}*/intMain () {intt,n,m; scanf ("%d",&t);  while(t--) {scanf ("%d%d",&n,&m);  for(intI=1; i<=n; i++) scanf ("%d",&Num[i]); Build (1N1); //cout<<tree[9].nm<<endl; //cout<< "*" <<endl;         for(intI=0; i<m; i++)        {            Charstr[2]; intb; scanf ("%s%d%d",str,&a,&b); if(str[0]=='U') Update (a+1B1N1); Else if(str[0]=='Q') printf ("%d\n", query (A +1, B +1,1N1)); }    }    return 0;}

Hdu_3308_ Segment Tree _ interval Merging

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.