Cdoj 1104 the intersection of the sub-columns of a two-series query interval is less than a number of the Chairmen tree

to find the intersection of the child columns of the two series

Time Limit:1 Sec

Memory limit:256 MB

Topic Connection http://acm.uestc.edu.cn/#/problem/show/1104
Description

Give two Series A, B, the length is N1, N2, respectively. 保证A中每个元素互不相同，保证B中每个元素互不相同 Make Q-queries, find A[L1...R1] and b[l2 each time. R2] The intersection collection size is:

For example A = {1,2,3,4,5,6,7},b = {7,6,5,4,3,2,1}

Query a[2..4] and b[3..5].. A[2..4] = {2,3,4};b[3..5] = {5,4,3}, with an intersection of {3,4} and a size of 2.

Input

The first line enters N1, the second line enters the number of N1; Similarly, the third row enters N2, and the fourth line enters the N2 number.

Enter Q in line five. Next Q Line input L1, R1, L2, R2.

Guaranteed N1, N2, q in [1,2x5 ] within the scope of1≤l1≤r1≤n1 ，1≤l2≤r2≤n2 , and then count in the [ −9,9] in the range:

Output

Output Q line, indicating the answer.

Sample Input

7
1 2 3 4 5 6 7
7
7 6 5 4 3 2 1
1
2 4 3 5

Sample Output

2HINT

Test instructions

Exercises

Chairman Tree naked, the query interval is less than a number of how many will be good

Code:

//Qscqesze#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<bitset>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>#include<map>#include<stack>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 200500#defineMoD 1001#defineEPS 1e-9#definePi 3.1415926intNum;//const int INF=0X7FFFFFFF;Constll inf=999999999; inline ll read () {ll x=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;}//*************************************************************************************inta[maxn],b[maxn],lb,n,m;structsegmenttree{intL,r,s;} TREE[MAXN* -];intnode;intROOT[MAXN];intSearchintx) {    intL=0, r=Lb,mid;  while(l<r) {Mid=l+r+1>>1; if(b[mid]<=x) l=mid; Elser=mid-1; }    returnl;}intBuildintLintR) {    intk=++node; Tree[k].s=0; if(L==R)returnK; intM=l+r>>1; if(l<=m) tree[k].l=build (L,M); if(r>m) Tree[k].r=build (m+1, R); returnK;}intChangeintRtintLintRintx) {    intk=++node,root=K,mid; TREE[K]=Tree[rt]; Tree[k].s++;  while(l<r) {Mid=l+r>>1; if(x<=mid) {RT=TREE[RT].L; TREE[K].L=++node; K=node; TREE[K]=Tree[rt]; Tree[k].s++; R=mid; }        Else{RT=TREE[RT].R; TREE[K].R=++node; K=node; TREE[K]=Tree[rt]; Tree[k].s++; L=mid+1; }    }    returnRoot;}intQueryintLintRintLintRintx) {    intmid,ans=0;  while(l<r) {Mid=l+r>>1; if(mid>=x) {R=mid; L=TREE[L].L; R=TREE[R].L; }        Else{L=mid+1; Ans+=tree[tree[r].l].s-Tree[tree[l].l].s; L=TREE[L].R; R=TREE[R].R; }    }    if(l==r) {if(x<l)return 0; Else returnans+tree[r].s-Tree[l].s; }}int Get(intLintRintval) {Val=Search (val); if(val==0)return 0; returnQuery (root[l-1],root[r],1, Lb,val);}intA[MAXN];intMain () {intnn=read ();  for(intI=1; i<=nn;i++) A[i]=read (); Sort (A+1, a+nn+1); intn=read ();  for(intI=1; i<=n;i++)    {        intx=read (); intL =1, r =nn;  while(l<=r) {intMid = (l+r) >>1; if(a[mid]==x) {A[i]=x; Break;} if(a[mid]<x) l=mid+1; Elser = mid-1; }        if(a[i]==0) a[i]=inf; } lb=1;  for(intI=1; i<=n;i++) B[i]=A[i]; LB=1; Sort (b+1, B +1+N);  for(intI=2; i<=n;i++)        if(b[i]!=b[lb]) b[++lb]=C[i]; Node=0; root[0]=build (1, LB);  for(intI=1; i<=n;i++) Root[i]=change (root[i-1],1, Lb,search (A[i])); intm=read ();  for(intI=1; i<=m;i++)    {        intN1=read (), N2=read (), N3=read (), n4=read (); printf ("%d\n",Get(N3,N4,N2)-Get(n3,n4,n1-1)); }}

Cdoj 1104 the intersection of the sub-columns of a two-series query interval is less than a number of the Chairmen tree