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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).
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)
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.
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
Case #1: 13 7 3 6 9
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:\n",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\n",ans); 93 } 94 } 95 return 0; 96 }
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）
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原文地址：http://www.cnblogs.com/fraud/p/4397168.html
