Bzoj 1858: [Scoi2010] Sequence operation

1858: [Scoi2010] sequence operation time limit:10 Sec Memory limit:64 MB segment tree, for each interval need to maintain the left and right of the 1 and 0 consecutive number, and in op=4 special treatment. DESCRIPTIONLXHGWW recently received a 01 sequence, the sequence contains n number, these numbers are either 0, or 1, now for this sequence there are five transformation operations and inquiry operation: 0 A B to the [A, b] range of all the numbers into 0 1 a b [A, b] All the numbers in the interval become 1 2 A B all the numbers in the [A, a] range are reversed, in other words, turn all 0 into 1, turn all 1 into 0 3 a B and ask [A, b] the total number of 1 4 a B in the interval to ask [A, b] the maximum number of consecutive 1 for each inquiry operation, l XHGWW need to give answers, smart handlers, can you help him? Input data The first row consists of 2 numbers, N and M, respectively, indicating the length of the sequence and the number of operations the second line includes n number, indicating the initial state of the sequence next m, 3 numbers per line, OP, a, B, (0<=op<=4,0<=a<=b< N) indicates an operation marked op for interval [A, b] = "" "<=" "div=" ">output for each query operation, output one line, including 1 numbers, indicating its corresponding answer, sample Input10 10
0 0 0 1 1 0 1 0 1 1
1 0 2
3 0 5
2 2 2
4 0 4
0 3 6
2 3 7
4 2 8
1 0 5
0 5 6
3 3 9
Sample Output5
2
6
5

HINT

For 30% of data, 1<=n, m<=1000
For 100% of data, 1<=n, m<=100000

Source

#include <cstdio>#defineM 100010inlineintMaxintAintb) {returnA>b?a:b;}structtree{intL,R,S,LAZY,SUM,ML1,MR1,MM1,ML0,MR0,MM0;} tr[m*6];intA[m],n,m,x,y,v,op;inlineintRead () {inttmp=0; CharCh=GetChar ();  while(ch<'0'|| Ch>'9') ch=GetChar ();  while(ch>='0'&&ch<='9') {tmp=tmp*Ten+ch-'0'; ch=GetChar ();} returntmp;} InlinevoidPuintp) {    intp1=p<<1,p2=p<<1|1; Tr[p].sum=tr[p1].sum+tr[p2].sum; TR[P].MM1=tr[p1].mr1+TR[P2].ML1; TR[P].MM0=tr[p1].mr0+tr[p2].ml0; if(TR[P1].MR1==TR[P1].S) tr[p].ml1=tr[p1].mr1+TR[P2].ML1; Elsetr[p].ml1=TR[P1].ML1; if(TR[P1].MR0==TR[P1].S) tr[p].ml0=tr[p1].mr0+tr[p2].ml0; Elsetr[p].ml0=tr[p1].ml0; if(TR[P2].MR1==TR[P2].S) tr[p].mr1=tr[p2].mr1+TR[P1].MR1; Elsetr[p].mr1=TR[P2].MR1; if(TR[P2].MR0==TR[P2].S) tr[p].mr0=tr[p2].mr0+tr[p1].mr0; Elsetr[p].mr0=tr[p2].mr0; TR[P].MM1=Max (Tr[p].mm1,max (TR[P1].MM1,TR[P2].MM1)); TR[P].MM0=Max (Tr[p].mm0,max (TR[P1].MM0,TR[P2].MM0));} InlinevoidChangeintp) {    intt1=tr[p].ml0,t2=tr[p].mr0,t3=tr[p].mm0; Tr[p].ml0=TR[P].ML1; TR[P].ML1=T1; Tr[p].mr0=TR[P].MR1; TR[P].MR1=T2; TR[P].MM0=tr[p].mm1; TR[P].MM1=T3;} InlinevoidMakeintLintRintp) {TR[P].L=l,tr[p].r=r,tr[p].s=tr[p].r-tr[p].l+1, tr[p].lazy=-1; if(l==r) {tr[p].sum=A[l]; TR[P].ML1=tr[p].mm1=tr[p].mr1=A[l]; Tr[p].ml0=tr[p].mm0=tr[p].mr0=a[l]^1; return; }    intMid= (l+r) >>1; Make (L,mid,p<<1); Make (Mid+1,r,p<<1|1); PU (p);} InlinevoidPdintp) {    intlz=tr[p].lazy,p1=p<<1,p2=p<<1|1; if(lz==0) {Tr[p1].lazy=tr[p2].lazy=0; Tr[p1].sum=0; Tr[p2].sum=0; TR[P1].ML1=tr[p1].mr1=tr[p1].mm1=0; TR[P2].ML1=tr[p2].mr1=tr[p2].mm1=0; Tr[p1].ml0=tr[p1].mr0=tr[p1].mm0=Tr[p1].s; Tr[p2].ml0=tr[p2].mr0=tr[p2].mm0=Tr[p2].s; }    Else if(lz==1) {Tr[p1].lazy=tr[p2].lazy=1; Tr[p1].sum=Tr[p1].s; Tr[p2].sum=Tr[p2].s; TR[P1].ML1=tr[p1].mr1=tr[p1].mm1=Tr[p1].s; TR[P2].ML1=tr[p2].mr1=tr[p2].mm1=Tr[p2].s; Tr[p1].ml0=tr[p1].mr0=tr[p1].mm0=0; Tr[p2].ml0=tr[p2].mr0=tr[p2].mm0=0; }    Else if(lz==2)    {        if(tr[p1].lazy==1) tr[p1].lazy=0; Else if(tr[p1].lazy==0) tr[p1].lazy=1; Else if(tr[p1].lazy==2) tr[p1].lazy=-1; Elsetr[p1].lazy=2; if(tr[p2].lazy==1) tr[p2].lazy=0; Else if(tr[p2].lazy==0) tr[p2].lazy=1; Else if(tr[p2].lazy==2) tr[p2].lazy=-1; Elsetr[p2].lazy=2; Tr[p1].sum=tr[p1].s-tr[p1].sum; Tr[p2].sum=tr[p2].s-tr[p2].sum;    Change (p1), change (P2); } Tr[p].lazy=-1;} InlineintFind1 (intLintRintp)    {PD (P); if(TR[P].L==L&AMP;&AMP;TR[P].R==R)returntr[p].sum; intMid= (TR[P].L+TR[P].R) >>1; if(MID&GT;=R)returnFind1 (l,r,p<<1); Else if(mid<l)returnFind1 (l,r,p<<1|1); Else returnFind1 (l,mid,p<<1) +find1 (mid+1,r,p<<1|1);} InlineintFindl (intLintRintp)    {PD (P); if(Tr[p].ml1+l>r)returnr-l+1; Else returnTR[P].ML1;} InlineintFindr (intLintRintp)    {PD (P); if(L+tr[p].mr1>r)returnr-l+1; Else returnTR[P].MR1;} InlineintFind2 (intLintRintp)    {PD (P); if(TR[P].L==L&AMP;&AMP;TR[P].R==R)returnMax (Tr[p].ml1,max (TR[P].MM1,TR[P].MR1)); intMid= (TR[P].L+TR[P].R) >>1; if(MID&GT;=R)returnFind2 (l,r,p<<1); Else if(mid<l)returnFind2 (l,r,p<<1|1); Else returnMax (Max (Find2 (l,mid,p<<1), Find2 (mid+1,r,p<<1|1)), Findl (mid+1,r,p<<1|1) +findr (l,mid,p<<1));} InlinevoidxgintLintRintCintp)    {PD (P); if(tr[p].l==l&&tr[p].r==r) {Tr[p].lazy=C; if(c==0) {TR[P].ML1=tr[p].mr1=tr[p].mm1=0; Tr[p].ml0=tr[p].mr0=tr[p].mm0=Tr[p].s; Tr[p].sum=0; }        Else if(c==1) {tr[p].ml0=tr[p].mr0=tr[p].mm0=0; TR[P].ML1=tr[p].mr1=tr[p].mm1=Tr[p].s; Tr[p].sum=Tr[p].s; }        Else{tr[p].sum=tr[p].s-tr[p].sum;        Change (p); }        return; }    intMid= (TR[P].L+TR[P].R) >>1; if(mid>=r) XG (l,r,c,p<<1); Else if(mid<l) XG (l,r,c,p<<1|1); Else{xg (l,mid,c,p<<1); XG (Mid+1,r,c,p<<1|1); } PU (P);}intMain () {n=read (), m=read ();  for(intI=1; i<=n;i++) a[i]=read (); Make (1N1);  for(intI=0; i<m;i++) {op=read (); X=read (); y=read (); if(op<3) XG (x+1, y+1, OP,1); Else if(op==3) printf ("%d\n", Find1 (x+1, y+1,1)); Elseprintf"%d\n", Find2 (x+1, y+1,1)); }    return 0;}

Bzoj 1858: [Scoi2010] Sequence action