bzoj1251 sequence Terminator Splay

Wrote again splay feel more experience speed incredibly and stuck into the home.

The question to note is that the initial value should be-inf (MX and V)

#include <cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#defineRep (i,l,r) for (int i=l;i<r;i++)#defineCLR (a,x) memset (A,x,sizeof (a))using namespacestd;Const intmaxn=50050, maxm=100050, inf=0x3fffffff;intn,m;structnode*NULL,*pt;structnode{Node*ch[2]; intv,s,add,mx; BOOLRev; InlineintcmpintKConst{k-=ch[0]->s; if(k==1)return-1; returnk<=0?0:1; } Inlinevoidmaintain () {s=ch[0]->s+ch[1]->s+1; MX=max (V,max (ch[0]->mx,ch[1]->mx)); } Inlinevoidpushdown () {if(rev) {Rev=0; ch[0]->rev^=1; ch[1]->rev^=1; Swap (ch[0],ch[1]); }        if(add) {if(ch[0]!=NULL) {ch[0]->add+=add; ch[0]->v+=add; ch[0]->mx+=add; }            if(ch[1]!=NULL) {ch[1]->add+=add; ch[1]->mx+=add; ch[1]->v+=add; } Add=0; }}};node*NewNode () {pt->ch[0]=NULL; PT->ch[1]=NULL; PT->mx=0; PT->add=0; PT->v=0; PT->s=1; PT->rev=0; returnpt++;} Node*Root;node X[MAXN];voidRotate (Node*&o,intd) {Node*k=o->ch[d^1]; o->ch[d^1]=k->Ch[d]; K->ch[d]=o; o->maintain ();k->maintain (); o=K;}voidSplay (Node*&o,intk) {o-pushdown (); intD=o->CMP (k); if(d==-1)return; if(d==1) k-=o->ch[0]->s+1; Node*p=o->Ch[d]; P-pushdown (); intD2=p->CMP (k); intk2=d2?k-p->ch[0]->s-1: K; if(d2!=-1) {splay (P-ch[d2],k2); D==d2?rotate (o,d^1): Rotate (o->ch[d],d); } Rotate (O,d^1);} Node* Build (intLintR) {        if(L&GT;=R)return NULL; intMid= (l+r) >>1; Node*o=NewNode (); if(L<mid) o->ch[0]=build (L,mid); if(mid+1&LT;R) o->ch[1]=build (mid+1, R); o-maintain (); returno;}intRead () {CharC; intf=1, ans=0; C=GetChar ();  while(!IsDigit (c)) {        if(c=='-') f=-1; C=GetChar (); }     while(IsDigit (c)) {ans=ans*Ten+c-'0'; C=GetChar (); }    returnf*ans;}voidinit () {PT=x; NULL=NewNode (); NULL->v=-inf; NULL->mx=-inf; NULL->s=0; Root=build (0, n+2);} intMain () {n=read (), m=read ();    Init ();  while(m--){        intopt=read (); if(opt==1){            intL=read (), R=read (), delta=read ();            Splay (root,l); Splay (Root->ch[1],r+1-root->ch[0]->s); Root->ch[1]->ch[0]->add+=Delta; Root->ch[1]->ch[0]->v+=Delta; Root->ch[1]->ch[0]->mx+=Delta; }Else if(opt==2){            intL=read (), r=read (); if(L==R)Continue;            Splay (root,l); Splay (Root->ch[1],r+1-root->ch[0]->s); Root->ch[1]->ch[0]->rev^=1; }Else{            intL=read (), r=read ();            Splay (root,l); Splay (Root->ch[1],r+1-root->ch[0]->s); printf ("%d\n", root->ch[1]->ch[0]->MX); }    }    return 0;}

View Code 1251: Sequence terminator time limit:20 Sec Memory limit:162 MB
submit:2791 solved:1110
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There are many questions on the web, that is, given a sequence, you need to support several operations: A, B, C, D. Look at another question, and another sequence to support several operations: D, C, B, A. In particular, we have someone here, out of the mock test, incredibly also out of one such, really no technical content ... So I also out of a problem, I came out this purpose is to let everyone do this topic has a "library" can rely on, there is no other meaning. This problem is called sequence Terminator. "Problem description" given a sequence of length n, the element of each sequence is an integer (nonsense). To support the following three actions: 1. Add all the numbers in this range [L,r] to V. 2. Flip [l,r] this interval, e.g. 1 2 3 4 to 4 3 2 1. 3. Ask for the maximum value in this interval [l,r]. At the very beginning, all elements are 0.

Input

The first row of two integers n,m. M is the number of operations. The following m lines, with a maximum of four integers per line, are k,l,r,v. K is the number of operations, if not the 1th operation then there are only two digits after K.

Output

For each of the 3rd operations, give the correct answer.

Sample Input4 4
1 1 3 2
1 2 4-1
2 1 3
3 2 4

Sample Output2
"Data Range"
n<=50000,m<=100000.

HINT Source

Splay

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bzoj1251 sequence Terminator Splay