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.

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

 

 

划分树：

 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 }
View Code

 

主席树（循环）：

  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 }
View Code

 

主席树(递归）：

  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 }
View Code

 

 

 


 

posted on 2018-04-03 20:59  h_z_cong  阅读(170)  评论(0)  编辑  收藏  举报

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