POJ2104 K-th Number —— 静态区间第k小
题目链接:http://poj.org/problem?id=2104
K-th Number
Time Limit: 20000MS | Memory Limit: 65536K | |
Total Submissions: 64277 | Accepted: 22615 | |
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.
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).
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
划分树:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int mod = 1e9+7; 17 const int MAXN = 1e5+10; 18 19 int tree[20][MAXN]; 20 int sorted[MAXN]; 21 int toleft[20][MAXN]; 22 23 void build(int l, int r, int dep) 24 { 25 if(l==r) return; 26 int mid = (l+r)>>1; 27 int same = mid-l+1; 28 for(int i = l; i<=r; i++) 29 if(tree[dep][i]<sorted[mid]) 30 same--; 31 32 int lpos = l, rpos = mid+1; 33 for(int i = l; i<=r; i++) 34 { 35 if(tree[dep][i]<sorted[mid]) 36 tree[dep+1][lpos++] = tree[dep][i]; 37 else if(tree[dep][i]==sorted[mid] && same>0) 38 { 39 tree[dep+1][lpos++] = tree[dep][i]; 40 same--; 41 } 42 else 43 tree[dep+1][rpos++] = tree[dep][i]; 44 toleft[dep][i] = toleft[dep][l-1] + lpos - l; 45 } 46 47 build(l, mid, dep+1); 48 build(mid+1, r, dep+1); 49 } 50 51 int query(int L, int R, int l, int r, int dep, int k) 52 { 53 if(l==r) return tree[dep][l]; 54 int mid = (L+R)>>1; 55 int cnt = toleft[dep][r] - toleft[dep][l-1]; 56 57 if(cnt>=k) 58 { 59 int newl = L + toleft[dep][l-1] - toleft[dep][L-1]; 60 int newr = newl + cnt - 1; 61 return query(L, mid, newl, newr, dep+1, k); 62 } 63 else 64 { 65 int newr = r + toleft[dep][R] - toleft[dep][r]; 66 int newl = newr - (r-l-cnt); 67 return query(mid+1, R, newl, newr, dep+1, k-cnt); 68 } 69 } 70 71 int main() 72 { 73 int n, m; 74 while(scanf("%d%d",&n,&m)!=EOF) 75 { 76 memset(tree, 0, sizeof(tree)); 77 for(int i = 1; i<=n; i++) 78 { 79 scanf("%d",&tree[0][i]); 80 sorted[i] = tree[0][i]; 81 } 82 sort(sorted+1, sorted+1+n); 83 build(1, n, 0); 84 int s, t, k; 85 while(m--) 86 { 87 scanf("%d%d%d",&s,&t,&k); 88 printf("%d\n", query(1,n,s,t,0,k)); 89 } 90 } 91 return 0; 92 }
主席树(循环):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 1e5+10; 18 const int M = MAXN*30; 19 20 int n, q, m, tot; 21 int a[MAXN], t[MAXN]; 22 int T[MAXN], lson[M], rson[M], c[M]; 23 24 void Init_hash() 25 { 26 for(int i = 1; i<=n; i++) 27 t[i] = a[i]; 28 sort(t+1,t+1+n); 29 m = unique(t+1,t+1+n)-(t+1); 30 } 31 32 int hash(int x) 33 { 34 return lower_bound(t+1,t+1+m, x)-t; 35 } 36 37 int build(int l, int r) 38 { 39 int root = tot++; 40 c[root] = 0; 41 if(l!=r) 42 { 43 int mid = (l+r)>>1; 44 lson[root] = build(l,mid); 45 rson[root] = build(mid+1,r); 46 } 47 return root; 48 } 49 50 int update(int root, int pos, int val) 51 { 52 int newroot = tot++, tmp = newroot; 53 c[newroot] = c[root] + val; 54 int l = 1, r = m; 55 while(l<r) 56 { 57 int mid = (l+r)>>1; 58 if(pos<=mid) 59 { 60 lson[newroot] = tot++; rson[newroot] = rson[root]; 61 newroot = lson[newroot]; root = lson[root]; 62 r = mid; 63 } 64 else 65 { 66 rson[newroot] = tot++; lson[newroot] = lson[root]; 67 newroot = rson[newroot]; root = rson[root]; 68 l = mid + 1; 69 } 70 c[newroot] = c[root] + val; 71 } 72 return tmp; 73 } 74 75 int query(int left_root, int right_root, int k) 76 { 77 int l = 1, r = m; 78 while(l<r) 79 { 80 int mid = (l+r)>>1; 81 if(c[lson[left_root]]-c[lson[right_root]]>=k) 82 { 83 r = mid; 84 left_root = lson[left_root]; 85 right_root = lson[right_root]; 86 } 87 else 88 { 89 l = mid + 1; 90 k -= c[lson[left_root]]-c[lson[right_root]]; 91 left_root = rson[left_root]; 92 right_root = rson[right_root]; 93 } 94 } 95 return l; 96 } 97 98 int main() 99 { 100 while(scanf("%d%d",&n,&q)==2) 101 { 102 tot = 0; 103 for(int i = 1; i<=n; i++) 104 scanf("%d",&a[i]); 105 Init_hash(); 106 T[n+1] = build(1,m); 107 for(int i = n; i; i--) 108 { 109 int pos = hash(a[i]); 110 T[i] = update(T[i+1],pos,1); 111 } 112 while(q--) 113 { 114 int l, r, k; 115 scanf("%d%d%d",&l,&r,&k); 116 printf("%d\n", t[query(T[l],T[r+1],k)]); 117 } 118 } 119 }
主席树(递归):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <cmath> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const int INF = 2e9; 15 const LL LNF = 9e18; 16 const int MOD = 1e9+7; 17 const int MAXN = 1e5+10; 18 const int M = MAXN*30; 19 20 int n, q, m, tot; 21 int a[MAXN], t[MAXN]; 22 int T[MAXN], lson[M], rson[M], c[M]; 23 24 void Init_hash() 25 { 26 for(int i = 1; i<=n; i++) 27 t[i] = a[i]; 28 sort(t+1,t+1+n); 29 m = unique(t+1,t+1+n)-(t+1); 30 } 31 32 int hash(int x) 33 { 34 return lower_bound(t+1,t+1+m, x)-t; 35 } 36 37 int build(int l, int r) 38 { 39 int root = tot++; 40 c[root] = 0; 41 if(l!=r) 42 { 43 int mid = (l+r)>>1; 44 lson[root] = build(l,mid); 45 rson[root] = build(mid+1,r); 46 } 47 return root; 48 } 49 50 int update(int root, int l, int r, int pos, int val) 51 { 52 int newroot = tot++; 53 if(l==r) 54 { 55 c[newroot] = c[root] + val; 56 return newroot; 57 } 58 int mid = (l+r)>>1; 59 if(pos<=mid) 60 { 61 lson[newroot] = update(lson[root],l,mid,pos,val); 62 rson[newroot] = rson[root]; 63 } 64 else 65 { 66 rson[newroot] = update(rson[root],mid+1,r,pos,val); 67 lson[newroot] = lson[root]; 68 } 69 c[newroot] = c[lson[newroot]] + c[rson[newroot]]; 70 return newroot; 71 } 72 73 int query(int left_root, int right_root, int l, int r, int k) 74 { 75 if(l==r) return l; 76 77 int mid = (l+r)>>1; 78 if(c[lson[left_root]]-c[lson[right_root]]>=k) 79 return query(lson[left_root],lson[right_root],l,mid,k); 80 else 81 return query(rson[left_root],rson[right_root],mid+1,r,k-(c[lson[left_root]]-c[lson[right_root]])); 82 } 83 84 int main() 85 { 86 while(scanf("%d%d",&n,&q)!=EOF) 87 { 88 tot = 0; 89 for(int i = 1; i<=n; i++) 90 scanf("%d",&a[i]); 91 Init_hash(); 92 T[n+1] = build(1,m); 93 for(int i = n; i; i--) 94 { 95 int pos = hash(a[i]); 96 T[i] = update(T[i+1],1,m,pos,1); 97 } 98 while(q--) 99 { 100 int l, r, k; 101 scanf("%d%d%d",&l,&r,&k); 102 printf("%d\n", t[query(T[l],T[r+1],1,m,k)]); 103 } 104 } 105 }