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。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <cstdlib>
 7 #include <vector>
 8 using namespace std;
 9 #define w(i) T[i].w
10 #define ls(i) T[i].ls
11 #define rs(i) T[i].rs
12 #define MAXN 100010
13 int p[MAXN],a[MAXN],b[MAXN],root[MAXN];
14 struct node{
15     int ls,rs,w;
16     node(){ls=rs=w=0;}
17 }T[MAXN*20];
18 int tot=0;
19 void Insert(int &i,int l,int r,int x){
20     T[++tot]=T[i];
21     i=tot;
22     w(i)++;
23     if(l==r)return;
24     int mid=(l+r)>>1;
25     if(x<=mid)Insert(ls(i),l,mid,x);
26     else Insert(rs(i),mid+1,r,x);
27 }
28 int query(int lx,int rx,int l,int r,int k){
29     if(l==r)return l;
30     int ret=w(ls(rx))-w(ls(lx));
31     int mid=(l+r)>>1;
32     if(ret>=k)return query(ls(lx),ls(rx),l,mid,k);
33     else return query(rs(lx),rs(rx),mid+1,r,k-ret);
34 }
35 bool cmp(int i,int j){
36     return a[i]<a[j];
37 }
38 int main()
39 {
40     ios::sync_with_stdio(false);
41     tot=0;
42     int n,m;
43     int t;
44     int cas=1;
45     scanf("%d",&t);
46     while(t--){
47         scanf("%d%d",&n,&m);
48         for(int i=1;i<=n;i++){
49             scanf("%d",a+i);
50             p[i]=i;
51         }
52         tot=0;
53         root[0]=0;
54         sort(p+1,p+n+1,cmp);
55         for(int i=1;i<=n;i++)b[p[i]]=i;
56         for(int i=1;i<=n;i++){
57             root[i]=root[i-1];
58             Insert(root[i],1,n,b[i]);
59         }
60         printf("Case #%d:
",cas++);
61         while(m--){
62             int l,r,x,y;
63             scanf("%d%d%d%d",&l,&r,&x,&y);
64             int lx=1,rx=r-l+1;
65             int fx=-1;
66             int ans=0;
67             int flag =0;
68             int tmpx;
69             while(lx<=rx){
70                 int mid = (lx+rx)>>1;
71                 tmpx=a[p[query(root[l-1],root[r],1,n,mid)]];
72                 if(tmpx<=y){
73                     fx=mid;
74                     lx=mid+1;
75                 }else rx=mid-1;
76             }
77             if(fx==-1)flag=1;
78             else ans+=fx;
79             lx=1,rx=r-l+1;
80             fx=-1;
81             while(lx<=rx){
82                 int mid = (lx+rx)>>1;
83                 tmpx=a[p[query(root[l-1],root[r],1,n,mid)]];
84                 if(tmpx>=x){
85                     fx=mid;
86                     rx=mid-1;
87                 }else lx=mid+1;
88             }
89             if(fx==-1)flag=1;
90             else ans=ans-fx+1;
91             if(flag)ans=0;
92             printf("%d
",ans);
93         }
94     }
95     return 0;
96 }