K-th Number Time Limit: 20000MS Memory Limit: 65536K Total Submissions: 57427 Accepted: 19856 Case Time Limit: 2000MS Description You are working for ...
K-th Number
| Time Limit: 20000MS | Memory Limit: 65536K | |
| Total Submissions: 57427 | Accepted: 19856 | |
| Case Time Limit: 2000MS | ||
Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.Sample Input
7 3 1 5 2 6 3 7 4 2 5 3 4 4 1 1 7 3
Sample Output
5 6 3
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.Source
Northeastern Europe 2004, Northern Subregion 主席樹的模板題。1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 const int MAXN=10000001; 9 void read(int &n) 10 { 11 char c='+';int x=0;bool flag=0; 12 while(c<'0'||c>'9') 13 {c=getchar();if(c=='-')flag=1;} 14 while(c>='0'&&c<='9') 15 {x=x*10+c-48,c=getchar();} 16 flag==1?n=-x:n=x; 17 } 18 int n,m; 19 int a[MAXN]; 20 int hash[MAXN]; 21 int root[MAXN/2]; 22 int now,tot;// 所有節點的總出現次數 23 int ls[MAXN],rs[MAXN],cnt[MAXN]; 24 void build(int &cur,int l,int r) 25 { 26 cur=tot++;// 總的節點數量 27 cnt[cur]=0; 28 if(l!=r) 29 { 30 int mid=(l+r)/2; 31 build(ls[cur],l,mid); 32 build(rs[cur],mid+1,r); 33 } 34 } 35 void update(int pre,int pos,int &cur,int l,int r) 36 { 37 cur=tot++; 38 cnt[cur]=cnt[pre]+1; 39 ls[cur]=ls[pre];rs[cur]=rs[pre]; 40 if(l==r) 41 return ; 42 int mid=(l+r)/2; 43 if(pos<=mid) 44 update(ls[pre],pos,ls[cur],l,mid); 45 else 46 update(rs[pre],pos,rs[cur],mid+1,r); 47 } 48 int query(int lt,int rt,int l,int r,int k) 49 { 50 if(l==r) 51 return l; 52 int now=cnt[ls[rt]]-cnt[ls[lt]]; 53 int mid=(l+r)/2; 54 if(k<=now) 55 return query(ls[lt],ls[rt],l,mid,k); 56 else 57 return query(rs[lt],rs[rt],mid+1,r,k-now); 58 } 59 int main() 60 { 61 while(scanf("%d%d",&n,&m)==2) 62 { 63 for(int i=1;i<=n;i++) 64 { 65 read(a[i]); 66 hash[i]=a[i]; 67 } 68 sort(hash+1,hash+n+1); 69 int size=unique(hash+1,hash+n+1)-hash-1; 70 for(int i=1;i<=n;i++) 71 a[i]=lower_bound(hash+1,hash+size+1,a[i])-hash;// 排序+離散化 72 73 tot=0; 74 build(root[0],1,size); 75 76 for(int i=1;i<=n;i++) 77 update(root[i-1],a[i],root[i],1,size);// 建出所有的樹 78 79 for(int i=1;i<=m;i++) 80 { 81 int x,y,z; 82 read(x);read(y);read(z); 83 printf("%d\n",hash[query(root[x-1],root[y],1,size,z)]); 84 } // 查詢 85 } 86 87 return 0; 88 }
