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
[ Submit][ Status][ Discuss]

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、

 

 

  1 #include<cstdio> 

  2 #include<iostream> 

  3 #include<cstdlib> 

  4 #include<algorithm> 

  5 #include<cstring> 

  6 #define N 100010 

  7 using namespace std; 

  8   

  9 int toleft[30][N]; 

 10 int Tree[30][N]; 

 11 int Sort[N]; 

 12   

 13   

 14   

 15 void build(int l,int r,int dep) 

 16 { 

 17     int mid=(l+r)/2,sum=mid-l+1,lpos,rpos,i; 

 18     if(l==r) return ; 

 19     for(i=l;i<=r;i++) 

 20         if(Tree[dep][i]<Sort[mid]) sum--; 

 21     lpos=l; 

 22     rpos=mid+1; 

 23     for(i=l;i<=r;i++) 

 24     { 

 25         if(Tree[dep][i]<Sort[mid]) 

 26         { 

 27             Tree[dep+1][lpos++]=Tree[dep][i]; 

 28         } 

 29         else if(Tree[dep][i]==Sort[mid] && sum>0) 

 30         { 

 31             Tree[dep+1][lpos++]=Tree[dep][i]; 

 32             sum--; 

 33         } 

 34         else

 35         { 

 36             Tree[dep+1][rpos++]=Tree[dep][i]; 

 37         } 

 38         toleft[dep][i]=toleft[dep][l-1]+lpos-l; 

 39     } 

 40     build(l,mid,dep+1); 

 41     build(mid+1,r,dep+1); 

 42 } 

 43   

 44 int query(int L,int R,int l,int r,int dep,int k) 

 45 { 

 46     int mid=(L+R)/2,newl,newr,cur; 

 47     if(l==r) return Tree[dep][l]; 

 48     cur=toleft[dep][r]-toleft[dep][l-1]; 

 49     if(cur>=k) 

 50     { 

 51         newl=L+toleft[dep][l-1]-toleft[dep][L-1]; 

 52         newr=newl+cur-1; 

 53         return query(L,mid,newl,newr,dep+1,k); 

 54     } 

 55     else

 56     { 

 57         newr=r+(toleft[dep][R]-toleft[dep][r]); 

 58         newl=newr-(r-l-cur); 

 59         return query(mid+1,R,newl,newr,dep+1,k-cur); 

 60     } 

 61 } 

 62 int EF1(int l,int r,int num,int n) //枚举下届

 63 { 

 64     int low=1,right=r-l+1,mid,tmp; 

 65     mid=(low+right)/2; 

 66     while(low<right) 

 67     { 

 68         tmp=query(1,n,l,r,0,mid); 

 69         if(tmp>=num) 

 70         right=mid-1; 

 71         else low=mid; //！！！！

 72         mid=(low+right+1)/2; //!!! 加了一个1。

 73     } 

 74     return low; 

 75   

 76 } 

 77   

 78 int EF2(int l,int r,int num,int n) 

 79 { 

 80     int low=1,right=r-l+1,mid,tmp; 

 81     mid=(low+right)/2; 

 82     while(low<right) 

 83     { 

 84         tmp=query(1,n,l,r,0,mid); 

 85         if(tmp>num) 

 86         right=mid; //!!!

 87         else low=mid+1; 

 88         mid=(low+right)/2; //木有加1

 89     } 

 90     return low; 

 91 } 

 92   

 93 int main() 

 94 { 

 95     int T,n,m,i,l,r,k,a,b,hxl,tom,qq; 

 96     scanf("%d",&T); 

 97     { 

 98         for(qq=1;qq<=T;qq++) 

 99         { 

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

101             memset(Tree,0,sizeof(Tree)); 

102             for(i=1;i<=n;i++) 

103             { 

104                 scanf("%d",&Tree[0][i]); 

105                 Sort[i]=Tree[0][i]; 

106             } 

107             sort(Sort+1,Sort+1+n); 

108             build(1,n,0); 

109             printf("Case #%d:\n",qq); 

110             while(m--) 

111             { 

112                 scanf("%d%d%d%d",&l,&r,&a,&b); 

113                 hxl=query(1,n,l,r,0,1); 

114                 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

115                 else

116                 { 

117                     hxl=EF1(l,r,a,n); 

118                     hxl++; 

119                 } 

120                 tom=query(1,n,l,r,0,r-l+1); 

121                 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

122                 else

123                 { 

124                     tom=EF2(l,r,b,n); 

125                     tom--; 

126                 } 

127                 hxl=tom-hxl+1; 

128                 printf("%d\n",hxl); 

129             } 

130         } 

131     } 

132     return 0; 

133 }