K-th Number
| Time Limit: 20000MS | Memory Limit: 65536K | |
| Total Submissions: 59327 | Accepted: 20660 | |
| 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
//题意: n 个数，m 次询问，求 l,r,中第 k 小的数是什么
题解:有很多方法可以写，线段树配上归并，我用来当做主席树模板来做，学习！
1 # include <cstdio> 2 # include <cstring> 3 # include <cstdlib> 4 # include <iostream> 5 # include <vector> 6 # include <queue> 7 # include <stack> 8 # include <map> 9 # include <bitset> 10 # include <sstream> 11 # include <set> 12 # include <cmath> 13 # include <algorithm> 14 # pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 # define LL long long 17 # define pb push_back 18 # define pr pair 19 # define mkp make_pair 20 # define lowbit(x) ((x)&(-x)) 21 # define PI acos(-1.0) 22 # define INF 0x3f3f3f3f3f3f3f3f 23 # define eps 1e-8 24 # define MOD 1000000007 25 26 inline int scan() { 27 int x=0,f=1; char ch=getchar(); 28 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 29 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 30 return x*f; 31 } 32 inline void Out(int a) { 33 if(a<0) {putchar('-'); a=-a;} 34 if(a>=10) Out(a/10); 35 putchar(a%10+'0'); 36 } 37 # define MX 100005 38 /**************************/ 39 struct Node 40 { 41 int l,r; 42 int sum; 43 }tree[MX*40]; 44 45 int n,m; 46 int newn,tcnt; 47 int dat[MX]; 48 int root[MX]; 49 int cpy[MX]; 50 int getid(int x){return lower_bound(cpy+1,cpy+1+n,x)-cpy;} 51 52 int update(int l,int r,int y,int pos) 53 { 54 tcnt++; 55 tree[tcnt].sum = tree[y].sum+1; 56 int sav = tcnt; 57 58 if (l==r) return sav; 59 int mid = (l+r)/2; 60 if (pos<=mid) 61 { 62 tree[sav].l = update(l,mid,tree[y].l,pos); 63 tree[sav].r = tree[y].r; 64 } 65 else 66 { 67 tree[sav].l = tree[y].l; 68 tree[sav].r = update(mid+1,r,tree[y].r,pos); 69 } 70 return sav; 71 } 72 73 int inqy(int l,int r,int x,int y,int k) 74 { 75 if (l==r) return l; 76 int all =tree[ tree[y].l ].sum - tree[ tree[x].l ].sum; 77 int mid = (l+r)/2; 78 if (k<=all) return inqy(l,mid,tree[x].l,tree[y].l,k); 79 else return inqy(mid+1,r,tree[x].r,tree[y].r,k-all); 80 81 } 82 83 84 int main () 85 { 86 while (scanf("%d%d",&n,&m)!=EOF) 87 { 88 for (int i=1;i<=n;i++) 89 { 90 dat[i] = scan(); 91 cpy[i] = dat[i]; 92 } 93 sort(cpy+1,cpy+1+n); 94 newn = unique(cpy+1,cpy+1+n)-cpy; 95 tcnt=0; 96 for (int i=1;i<=n;i++) root[i]=update(1,n,root[i-1],getid(dat[i])); 97 while (m--) 98 { 99 int a = scan(); 100 int b = scan(); 101 int k = scan(); 102 printf("%d ",cpy[inqy(1,n,root[a-1],root[b],k)]); 103 } 104 } 105 return 0; 106 }