山东第四届省赛： Boring Counting 线段树

http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=3237

Problem H:Boring Counting

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 6  Solved: 2
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Description

In this problem you are given a number sequence P consisting of N integer and P_i is the i^th 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 P_i is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of P_i (L <= i <= R,  A <= P_i <= 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 P_i(1 <= P_i <= 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

山东省第四届ACM程序设计大赛2013.6.9

比赛的时候，没有头绪，今天别人说起划分树，想了一下，还真是可以做的。

但是要一点变形，这样的变形，让我已经看不清 划分树的影子了。

划分树----求区间第K 小/大 值

这道题是求区间求区间[l,r]中 满足  A<=xi<=B; xi属于[l,r]。求xi的个数。

如何转化的呢？ 

                     枚举区间第k小数==A。此时能求出一个值 hxl

                     枚举区间第K小数==B，此时得到一个值   tom

                     显然满足tom>=hxl。

                     那在这范围的数，是不是就满足了  A<=xi<=B  !~~

这么一说，其实好简单。为什么没有想到呢？  划分树做的时候是求 第 k 小的是什么什么数字，现在换了回来而已，

求的是某个数字是区间里面第几小。

其实有个问题就又来了，A,B在区间[l,r]未必存在，这该怎么办？？？？？

其实上面的说法，不严谨，（求的是某个数字是区间里面第几小。）应该说是求A的下届，B的上届值。

二分枚举，平均的时间会更小。在求下届和上届的过程中也要注意一些细节的问题、

#include<cstdio> 
#include<iostream> 
#include<cstdlib> 
#include<algorithm> 
#include<cstring> 
#define N 100010 
using namespace std; 
  
int toleft[30][N]; 
int Tree[30][N]; 
int Sort[N]; 
  
  
  
void build(int l,int r,int dep) 
{ 
    int mid=(l+r)/2,sum=mid-l+1,lpos,rpos,i; 
    if(l==r) return ; 
    for(i=l;i<=r;i++) 
        if(Tree[dep][i]<Sort[mid]) sum--; 
    lpos=l; 
    rpos=mid+1; 
    for(i=l;i<=r;i++) 
    { 
        if(Tree[dep][i]<Sort[mid]) 
        { 
            Tree[dep+1][lpos++]=Tree[dep][i]; 
        } 
        else if(Tree[dep][i]==Sort[mid] && sum>0) 
        { 
            Tree[dep+1][lpos++]=Tree[dep][i]; 
            sum--; 
        } 
        else
        { 
            Tree[dep+1][rpos++]=Tree[dep][i]; 
        } 
        toleft[dep][i]=toleft[dep][l-1]+lpos-l; 
    } 
    build(l,mid,dep+1); 
    build(mid+1,r,dep+1); 
} 
  
int query(int L,int R,int l,int r,int dep,int k) 
{ 
    int mid=(L+R)/2,newl,newr,cur; 
    if(l==r) return Tree[dep][l]; 
    cur=toleft[dep][r]-toleft[dep][l-1]; 
    if(cur>=k) 
    { 
        newl=L+toleft[dep][l-1]-toleft[dep][L-1]; 
        newr=newl+cur-1; 
        return query(L,mid,newl,newr,dep+1,k); 
    } 
    else
    { 
        newr=r+(toleft[dep][R]-toleft[dep][r]); 
        newl=newr-(r-l-cur); 
        return query(mid+1,R,newl,newr,dep+1,k-cur); 
    } 
} 
int EF1(int l,int r,int num,int n) //枚举下届
{ 
    int low=1,right=r-l+1,mid,tmp; 
    mid=(low+right)/2; 
    while(low<right) 
    { 
        tmp=query(1,n,l,r,0,mid); 
        if(tmp>=num) 
        right=mid-1; 
        else low=mid; //！！！！
        mid=(low+right+1)/2; //!!! 加了一个1。
    } 
    return low; 
  
} 
  
int EF2(int l,int r,int num,int n) 
{ 
    int low=1,right=r-l+1,mid,tmp; 
    mid=(low+right)/2; 
    while(low<right) 
    { 
        tmp=query(1,n,l,r,0,mid); 
        if(tmp>num) 
        right=mid; //!!!
        else low=mid+1; 
        mid=(low+right)/2; //木有加1
    } 
    return low; 
} 
  
int main() 
{ 
    int T,n,m,i,l,r,k,a,b,hxl,tom,qq; 
    scanf("%d",&T); 
    { 
        for(qq=1;qq<=T;qq++) 
        { 
            scanf("%d%d",&n,&m); 
            memset(Tree,0,sizeof(Tree)); 
            for(i=1;i<=n;i++) 
            { 
                scanf("%d",&Tree[0][i]); 
                Sort[i]=Tree[0][i]; 
            } 
            sort(Sort+1,Sort+1+n); 
            build(1,n,0); 
            printf("Case #%d:
",qq); 
            while(m--) 
            { 
                scanf("%d%d%d%d",&l,&r,&a,&b); 
                hxl=query(1,n,l,r,0,1); 
                if(a<=hxl) hxl=1; //对临界条件的判断。如果A就是最小的，木有下届
                else
                { 
                    hxl=EF1(l,r,a,n); 
                    hxl++; 
                } 
                tom=query(1,n,l,r,0,r-l+1); 
                if(b>=tom) tom=r-l+1; //如果B是该区间[l,r]最大的，显然木有上届
                else
                { 
                    tom=EF2(l,r,b,n); 
                    tom--; 
                } 
                hxl=tom-hxl+1; 
                printf("%d
",hxl); 
            } 
        } 
    } 
    return 0; 
}