zoj2112 主席树动态第k大 （主席树&&树状数组）

Dynamic Rankings

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.

Your task is to write a program for this computer, which

- Reads N numbers from the input (1 <= N <= 50,000)

-
Processes M instructions of the input (1 <= M <= 10,000). These
instructions include querying the k-th smallest number of a[i], a[i+1],
..., a[j] and change some a[i] to t.

Input

The first line of the input is a
single number X (0 < X <= 4), the number of the test cases of the
input. Then X blocks each represent a single test case.

The first
line of each block contains two integers N and M, representing N
numbers and M instruction. It is followed by N lines. The (i+1)-th line
represents the number a[i]. Then M lines that is in the following format

Q i j k or
C i t

It
represents to query the k-th number of a[i], a[i+1], ..., a[j] and
change some a[i] to t, respectively. It is guaranteed that at any time
of the operation. Any number a[i] is a non-negative integer that is less
than 1,000,000,000.

There're NO breakline between two continuous test cases.

Output

For each querying operation,
output one integer to represent the result. (i.e. the k-th smallest
number of a[i], a[i+1],..., a[j])

There're NO breakline between two continuous test cases.

Sample Input

2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3

Sample Output

3
6
3
6

主席树太神了。

这题是动态第k大。

如果是不修改，直接主席树就可以了。

要修改要套如树状数组求和。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-9-8 8:53:54
 File Name     :F:\2013ACM练习\专题学习\主席树\ZOJ2112.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 const int MAXN = ;
 const int M = ;
 int n,q,m,tot;
 int a[MAXN], t[MAXN];
 int T[MAXN], lson[M], rson[M],c[M];
 int S[MAXN];

 struct Query
 {
     int kind;
     int l,r,k;
 }query[];

 void Init_hash(int k)
 {
     sort(t,t+k);
     m = unique(t,t+k) - t;
 }
 int hash(int x)
 {
     return lower_bound(t,t+m,x)-t;
 }
 int build(int l,int r)
 {
     int root = tot++;
     c[root] = ;
     if(l != r)
     {
         int mid = (l+r)/;
         lson[root] = build(l,mid);
         rson[root] = build(mid+,r);
     }
     return root;
 }

 int Insert(int root,int pos,int val)
 {
     int newroot = tot++, tmp = newroot;
     int l = , r = m-;
     c[newroot] = c[root] + val;
     while(l < r)
     {
         int mid = (l+r)>>;
         if(pos <= mid)
         {
             lson[newroot] = tot++; rson[newroot] = rson[root];
             newroot = lson[newroot]; root = lson[root];
             r = mid;
         }
         else
         {
             rson[newroot] = tot++; lson[newroot] = lson[root];
             newroot = rson[newroot]; root = rson[root];
             l = mid+;
         }
         c[newroot] = c[root] + val;
     }
     return tmp;
 }

 int lowbit(int x)
 {
     return x&(-x);
 }
 int use[MAXN];
 void add(int x,int pos,int val)
 {
     while(x <= n)
     {
         S[x] = Insert(S[x],pos,val);
         x += lowbit(x);
     }
 }
 int sum(int x)
 {
     int ret = ;
     while(x > )
     {
         ret += c[lson[use[x]]];
         x -= lowbit(x);
     }
     return ret;
 }
 int Query(int left,int right,int k)
 {
     int left_root = T[left-];
     int right_root = T[right];
     int l = , r = m-;
     for(int i = left-;i;i -= lowbit(i)) use[i] = S[i];
     for(int i = right;i ;i -= lowbit(i)) use[i] = S[i];
     while(l < r)
     {
         int mid = (l+r)/;
         int tmp = sum(right) - sum(left-) + c[lson[right_root]] - c[lson[left_root]];
         if(tmp >= k)
         {
             r = mid;
             for(int i = left-; i ;i -= lowbit(i))
                 use[i] = lson[use[i]];
             for(int i = right; i; i -= lowbit(i))
                 use[i] = lson[use[i]];
             left_root = lson[left_root];
             right_root = lson[right_root];
         }
         else
         {
             l = mid+;
             k -= tmp;
             for(int i = left-; i;i -= lowbit(i))
                 use[i] = rson[use[i]];
             for(int i = right;i ;i -= lowbit(i))
                 use[i] = rson[use[i]];
             left_root = rson[left_root];
             right_root = rson[right_root];
         }
     }
     return l;
 }
 void Modify(int x,int p,int d)
 {
     while(x <= n)
     {
         S[x] = Insert(S[x],p,d);
         x += lowbit(x);
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int Tcase;
     scanf("%d",&Tcase);
     while(Tcase--)
     {
         scanf("%d%d",&n,&q);
         tot = ;
         m = ;
         for(int i = ;i <= n;i++)
         {
             scanf("%d",&a[i]);
             t[m++] = a[i];
         }
         char op[];
         for(int i = ;i < q;i++)
         {
             scanf("%s",op);
             if(op[] == 'Q')
             {
                 query[i].kind = ;
                 scanf("%d%d%d",&query[i].l,&query[i].r,&query[i].k);
             }
             else
             {
                 query[i].kind = ;
                 scanf("%d%d",&query[i].l,&query[i].r);
                 t[m++] = query[i].r;
             }
         }
         Init_hash(m);
         T[] = build(,m-);
         for(int i = ;i <= n;i++)
             T[i] = Insert(T[i-],hash(a[i]),);
         for(int i = ;i <= n;i++)
             S[i] = T[];
         for(int i = ;i < q;i++)
         {
             if(query[i].kind == )
                 printf("%d\n",t[Query(query[i].l,query[i].r,query[i].k)]);
             else
             {
                 Modify(query[i].l,hash(a[query[i].l]),-);
                 Modify(query[i].l,hash(query[i].r),);
                 a[query[i].l] = query[i].r;
             }
         }
     }
     return ;
 }