13年山东省赛 Boring Counting（离线树状数组or主席树+二分or划分树+二分）

时间：2023-03-08 20:44:54

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2224: Boring Counting

Time Limit: 3 Sec  Memory Limit: 128 MB

Description

In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you 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 and 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

In the first line there is an integer T (1 < T <= 50), indicates the number of test cases. 
       For each case, the first line contains two 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 are M lines, each line contains four number L, R, A, B(1 <= L, R <= n, 1 <= A, B <= 10^9)

Output

For each case, at first output a line ‘Case #c:’, c is the case number start from 1. Then for each query output a line contains the answer.

Sample Input

1

13 5

6 9 5 2 3 6 8 7 3 2 5 1 4

1 13 1 10

1 13 3 6

3 6 3 6

2 8 2 8

1 9 1 9

Sample Output

Case #1:

13

7

3

6

9

HINT

Source

题意：求静态的区间[l,r]介于[A,B]的数的个数

这题其实是一道离线树状数组的水题，估计是因为我并不是很具备离线的思维吧，一般的离线都想不到。

好在主席树也能够完成这个操作，并且也只花了20分钟不到就1A了。

二分k的值，然后判断利用第k大来判断出A，B分别是第几大。然后随便搞。

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #include <cmath>

 #include <cstdlib>

 #include <vector>

 using namespace std;

 #define w(i) T[i].w

 #define ls(i) T[i].ls

 #define rs(i) T[i].rs

 #define MAXN 100010

 int p[MAXN],a[MAXN],b[MAXN],root[MAXN];

 struct node{

     int ls,rs,w;

     node(){ls=rs=w=;}

 }T[MAXN*];

 int tot=;

 void Insert(int &i,int l,int r,int x){

     T[++tot]=T[i];

     i=tot;

     w(i)++;

     if(l==r)return;

     int mid=(l+r)>>;

     if(x<=mid)Insert(ls(i),l,mid,x);

     else Insert(rs(i),mid+,r,x);

 }

 int query(int lx,int rx,int l,int r,int k){

     if(l==r)return l;

     int ret=w(ls(rx))-w(ls(lx));

     int mid=(l+r)>>;

     if(ret>=k)return query(ls(lx),ls(rx),l,mid,k);

     else return query(rs(lx),rs(rx),mid+,r,k-ret);

 }

 bool cmp(int i,int j){

     return a[i]<a[j];

 }

 int main()

 {

     ios::sync_with_stdio(false);

     tot=;

     int n,m;

     int t;

     int cas=;

     scanf("%d",&t);

     while(t--){

         scanf("%d%d",&n,&m);

         for(int i=;i<=n;i++){

             scanf("%d",a+i);

             p[i]=i;

         }

         tot=;

         root[]=;

         sort(p+,p+n+,cmp);

         for(int i=;i<=n;i++)b[p[i]]=i;

         for(int i=;i<=n;i++){

             root[i]=root[i-];

             Insert(root[i],,n,b[i]);

         }

         printf("Case #%d:\n",cas++);

         while(m--){

             int l,r,x,y;

             scanf("%d%d%d%d",&l,&r,&x,&y);

             int lx=,rx=r-l+;

             int fx=-;

             int ans=;

             int flag =;

             int tmpx;

             while(lx<=rx){

                 int mid = (lx+rx)>>;

                 tmpx=a[p[query(root[l-],root[r],,n,mid)]];

                 if(tmpx<=y){

                     fx=mid;

                     lx=mid+;

                 }else rx=mid-;

             }

             if(fx==-)flag=;

             else ans+=fx;

             lx=,rx=r-l+;

             fx=-;

             while(lx<=rx){

                 int mid = (lx+rx)>>;

                 tmpx=a[p[query(root[l-],root[r],,n,mid)]];

                 if(tmpx>=x){

                     fx=mid;

                     rx=mid-;

                 }else lx=mid+;

             }

             if(fx==-)flag=;

             else ans=ans-fx+;

             if(flag)ans=;

             printf("%d\n",ans);

         }

     }

     return ;

 }