hdu5737 (2016 Multi-school League 2nd game D)

Test instructions: to 2 sets of data A and B arrays, each with 2 operations: (+,L,R,X) The A array of the first to the first r elements are all x, (?, l,r) query [l,r] between the location of the a[i]>=b[i] (i>=l && i<= R) and count the number of these locations

have been thinking for a long time, did not think of any effective solutions, until the solution to see it, the original line of tree set balance tree also has this situation: inside is actually not a balance tree, just ordered table.

And then the problem is converted to the number of interval lookups corresponding to the ranking

Since this problem does not have to modify 2 arrays, of which 1 B-tree can be used as a fixed segment tree set ordered table to save time, the other 1 table a tree is simply using a line segment tree method first modified, and then update the corresponding B-tree node ranking

Find rankings here if all LOGN lookups are due to the constant too large direct card, note that each node contains a sequence table and contains a relationship up and down

Then we can update the parent node rank in the B-tree bottom-up corresponding to the ranking of the left and right sub-tree, using the method of merge sorting, Occupy space only O (Nlogn), Time is O (NLOGN)

By the way, a tree that will change the 1 nodes of the B-tree to find out the ranking, the modification of the root node corresponding to the location, and then according to the location of the sub-tree table one side downward update to the sub-tree corresponding position

#include <stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<Set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>using namespacestd;typedef __int64 ll;intMaxintAintb) {returnA>b?a:b;}intMinintAintb) {returnA<b?a:b;}intCNT;Const intn=100010, m=262150, e=1768950;intN,m,i,a[n],b[n],x,l,r;intst[m],en[m],v[m],tag[m],pl[e],pr[e],pool[e],cur;ll ans,sum;voidBuildintXintLintR) {Tag[x]=-1; if(l==R) {St[x]=cur+1; pool[++cur]=B[l]; EN[X]=cur; V[X]= (a[l]>=B[l]); return; }    intMid= (l+r) >>1); Build (x<<1, L,mid); Build ((x<<1)|1, mid+1, R); V[X]=v[x<<1]+v[(x<<1)|1]; intal=st[x<<1],ar=en[x<<1],bl=st[(x<<1)|1],br=en[(x<<1)|1]; ST[X]=cur+1;  while(AL&LT;=AR&AMP;&AMP;BL&LT;=BR) pool[++cur]=pool[al]<pool[bl]?pool[al++]:p ool[bl++];  while(Al<=ar) pool[++cur]=pool[al++];  while(BL&LT;=BR) pool[++cur]=pool[bl++]; EN[X]=cur; Al=st[x<<1],bl=st[x<<1|1];  for(inti=st[x];i<=cur;i++)    {         while(Al<=ar&&pool[al]<=pool[i]) al++;  while(Bl<=br&&pool[bl]<=pool[i]) bl++; Pl[i]=al-1, pr[i]=bl-1; if(pl[i]<st[x<<1]) pl[i]=0; if(Pr[i]<st[(x<<1)|1]) pr[i]=0; }}inlinevoidRANKPT (intXintp) {V[x]= (p?p-st[x]+1:0); TAG[X]=p;} InlinevoidPushdown (intx) {    if(tag[x]<0)return; intp=Tag[x]; RANKPT (x<<1, Pl[p]); Rankpt ((x<<1)|1, Pr[p]); TAG[X]=-1;}voidUpdateintXintAintBintp) {    if(L<=a && b<=r) {rankpt (x,p);return;}    Pushdown (x); intMid= (a+b) >>1; if(l<=mid) Update (x<<1, A,mid,pl[p]); if(r>mid) Update ((x<<1)|1, mid+1, B,pr[p]); V[X]=v[x<<1]+v[(x<<1)|1];}voidQueryintXintAintb) {    if(L<=a && b<=R) {ans+=V[x]; return;    } pushdown (x); intMid= (a+b) >>1); if(l<=mid) query (x<<1, A,mid); if(r>mid) query ((x<<1)|1, mid+1, B); V[X]=v[x<<1]+v[(x<<1)|1];} InlineintLowerintx) {    //Lower_bound (pool+st[1],pool+ed[1]+1,x);    intl=st[1],r=en[1],mid,t=0;  while(l<=R)if(Pool[mid= (L+r) >>1]<=x) l= (t=mid) +1; Elser=mid-1; returnt;}intSeeda, seedb, C = ~ (1<< to), MM = (1<< -)-1;intRndintLast ) {Seeda= (36969+ (Last >>3)) * (Seeda & MM) + (Seeda >> -); SeeDB= (18000+ (Last >>3)) * (SeeDB & MM) + (SeeDB >> -); return(C & (Seeda << -) (+ seedb))%1000000000;}intMain () {intT,ku; scanf ("%d",&t);  while(t--) {scanf ("%d%d%d%d",&n,&m,&seeda,&seedb);  for(i=1; i<=n;i++) scanf ("%d", A +i);  for(i=1; i<=n;i++) scanf ("%d", B +i); Ans=sum=cur=0; Build (1,1, N);  for(i=1; i<=m;i++) {L=rnd (ANS)%n+1, R=rnd (ANS)%n+1, Ku=rnd (ans) +1; intkkk=Lower (KU); if(L>r) l^=r^=l^=R; if((L+r+ku) &1) Update (1,1, N,lower (KU)); Else{ans=0; Query (1,1, N); Sum= (sum+ (ll) I*ans)%1000000007; }} printf ("%i64d\n", sum); }    return 0;}

View Code

hdu5737 (2016 Multi-school League 2nd game D)