13 Shandong Province Boring counting (offline tree array or chairman tree + dichotomy or partition tree + two points)

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2224:boring countingtime limit:3 Sec Memory limit:128 MB Description

In this problem given a number sequence P consisting of N integer and Pi was the ith element in the sequence. Now your task is to answer a list of queries, for each query, please tell us among [L, R], how many Pi is not less than a a nd not greater than B (l<= i <= R). In other words, your task is to count the number of pi (L <= i <= R, A <= Pi <= B).

Input

The first line there are an integer T (1 < T <=), indicates the number of test cases.
For each case, the first line contains numbers n and m (1 <= N, M <= 50000), the size of sequence P, the number of queries. The second line contains N numbers pi (1 <= pi <= 10^9), the number sequence P. Then there is M lines, each line contains four number L, R, A, B (1 <= L, R <= N, 1 <= A, B <= 10^9)

Output

For each case, an at first output of line ' case #c: ', C is the case number start from 1. Then for each query, output a line contains the answer.

Sample Input

113 56 9 5 2 3 6 8 7 3 2 5 1 41 13 1 101 13 3 63 6 3 62 8 2 81 9 1 9

Sample Output

Case #1:137369

HINT Source

2013 ACM College Student Program Design contest in Shandong Province

Test instructions: The number of static intervals [l,r] between [A, b]

This problem is actually an offline tree-like array of water problems, it is estimated that because I do not have the idea of offline, the general offline are unexpected.

Fortunately, the Chairman tree was able to do this, and it took only 20 minutes to 1 a.

The value of the binary K, and then judge the use of the K-large to determine the number of a or a, respectively, is the largest. and then casually.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <cmath>6#include <cstdlib>7#include <vector>8 using namespacestd;9 #defineW (i) T[I].WTen #defineLS (i) t[i].ls One #defineRS (i) t[i].rs A #defineMAXN 100010 - intP[MAXN],A[MAXN],B[MAXN],ROOT[MAXN]; - structnode{ the     intls,rs,w; -Node () {ls=rs=w=0;} -}t[maxn* -]; - inttot=0; + voidInsert (int&i,intLintRintx) { -t[++tot]=T[i]; +I=tot; AW (i) + +; at     if(L==R)return; -     intMid= (l+r) >>1; -     if(x<=mid) Insert (LS (i), l,mid,x); -     ElseInsert (RS (i), mid+1, r,x); - } - intQueryintLxintRxintLintRintk) { in     if(L==R)returnl; -     intRET=W (LS (RX))-W (LS (lx)); to     intMid= (l+r) >>1; +     if(ret>=k)returnquery (LS (lx), LS (RX), l,mid,k); -     Else returnQuery (RS (LX), RS (RX), mid+1, r,k-ret); the } * BOOLcmpintIintj) { $     returna[i]<A[j];Panax Notoginseng } - intMain () the { +Ios::sync_with_stdio (false); Atot=0; the     intn,m; +     intT; -     intcas=1; $scanf"%d",&t); $      while(t--){ -scanf"%d%d",&n,&m); -          for(intI=1; i<=n;i++){ thescanf"%d", A +i); -p[i]=i;Wuyi         } thetot=0; -root[0]=0; WuSort (p+1, p+n+1, CMP); -          for(intI=1; i<=n;i++) b[p[i]]=i; About          for(intI=1; i<=n;i++){ $root[i]=root[i-1]; -Insert (Root[i],1, N,b[i]); -         } -printf"Case #%d:\n", cas++); A          while(m--){ +             intl,r,x,y; thescanf"%d%d%d%d",&l,&r,&x,&y); -             intlx=1, rx=r-l+1; $             intfx=-1; the             intans=0; the             intFlag =0; the             inttmpx; the              while(lx<=Rx) { -                 intMid = (LX+RX) >>1; inTmpx=a[p[query (root[l-1],root[r],1, N,mid)]; the                 if(tmpx<=y) { thefx=mid; AboutLx=mid+1; the}Elserx=mid-1; the             } the             if(fx==-1) flag=1; +             Elseans+=FX; -lx=1, rx=r-l+1; thefx=-1;Bayi              while(lx<=Rx) { the                 intMid = (LX+RX) >>1; theTmpx=a[p[query (root[l-1],root[r],1, N,mid)]; -                 if(tmpx>=x) { -fx=mid; therx=mid-1; the}ElseLx=mid+1; the             } the             if(fx==-1) flag=1; -             Elseans=ans-fx+1; the             if(flag) ans=0; theprintf"%d\n", ans); the         }94     } the     return 0; the}

13 Shandong Province Boring counting (offline tree array or chairman tree + dichotomy or partition tree + two points)