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）

转载请注明出处： http://www.cnblogs.com/fraud/          ——by fraud

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

2013年山东省第四届ACM大学生程序设计竞赛

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 ;
 }