ZOJ 2112 Dynamic Rankings（主席树の动态kth）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2112

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.

题目大意：给n个数，有m个操作。修改某个数，或者询问一段区间的第k小值。

思路：首先要知道没有修改操作的区间第k大怎么用出席树写：POJ 2104 K-th Number（主席树——附讲解）

至于动态kth的解法可以去看CLJ的《可持久化数据结构研究》（反正我是看这个才懂的），然后在网上查一些资料，YY一下就可以了。

这里写的是树状数组套线段树+可持久化线段树的做法（因为内存不够用）。

简单的讲就是通过树状数组求和，维护前k个线段树的和。时间复杂度为O(nlogn+m(logn)^2)

另参考资料（里面有好几个link :)）：http://www.cnblogs.com/kuangbin/p/3308118.html

PS：ZOJ的指针似乎不是4个字节的。这里用指针就要MLE了（代码本来不是这么丑的啊T_T）。

代码（130MS）：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 typedef long long LL;

 const int MAXN = ;
 const int MAXM = ;
 const int MAXT = MAXM *  * ;

 struct Query {
     char op;
     int i, j, k;
     void read() {
         scanf(" %c", &op);
         if(op == 'Q') scanf("%d%d%d", &i, &j, &k);
         else scanf("%d%d", &i, &k);
     }
 } query[MAXM];
 int a[MAXN];
 int n, m, T;
 //hashmap
 int arr[MAXN + MAXM], total;

 void buildHash() {
     total = ;
     for(int i = ; i <= n; ++i) arr[total++] = a[i];
     for(int i = ; i <= m; ++i)
         if(query[i].op == 'C') arr[total++] = query[i].k;
     sort(arr, arr + total);
     total = unique(arr, arr + total) - arr;
 }

 int hash(int x) {
     return lower_bound(arr, arr + total, x) - arr;
 }
 //Chairman tree
 struct Node {
     int lson, rson, sum;
     void clear() {
         lson = rson = sum = ;
     }
 } tree[MAXT];
 int root[MAXN];
 int tcnt;

 void insert(int& rt, int l, int r, int pos) {
     tree[tcnt] = tree[rt]; rt = tcnt++;
     tree[rt].sum++;
     if(l < r) {
         int mid = (l + r) >> ;
         if(pos <= mid) insert(tree[rt].lson, l, mid, pos);
         else insert(tree[rt].rson, mid + , r, pos);
     }
 }

 void buildTree() {
     tcnt = ;
     for(int i = ; i <= n; ++i) {
         root[i] = root[i - ];
         insert(root[i], , total - , hash(a[i]));
     }
 }
 //Binary Indexed Trees
 int root2[MAXN];
 int roota[], rootb[];
 int cnta, cntb;

 void initBIT() {
     for(int i = ; i <= n; ++i) root2[i] = ;
 }

 inline int lowbit(int x) {
     return x & -x;
 }

 void insert(int& rt, int l, int r, int pos, int val) {
     if(rt == ) tree[rt = tcnt++].clear();
     if(l == r) {
         tree[rt].sum += val;
     } else {
         int mid = (l + r) >> ;
         if(pos <= mid) insert(tree[rt].lson, l, mid, pos, val);
         else insert(tree[rt].rson, mid + , r, pos, val);
         tree[rt].sum = tree[tree[rt].lson].sum + tree[tree[rt].rson].sum;
     }
 }

 int kth(int ra, int rb, int l, int r, int k) {
     if(l == r) {
         return l;
     } else {
         int sum = tree[tree[rb].lson].sum - tree[tree[ra].lson].sum, mid = (l + r) >> ;
         for(int i = ; i < cntb; ++i) sum += tree[tree[rootb[i]].lson].sum;
         for(int i = ; i < cnta; ++i) sum -= tree[tree[roota[i]].lson].sum;
         if(sum >= k) {
             for(int i = ; i < cntb; ++i) rootb[i] = tree[rootb[i]].lson;
             for(int i = ; i < cnta; ++i) roota[i] = tree[roota[i]].lson;
             return kth(tree[ra].lson, tree[rb].lson, l, mid, k);
         } else {
             for(int i = ; i < cntb; ++i) rootb[i] = tree[rootb[i]].rson;
             for(int i = ; i < cnta; ++i) roota[i] = tree[roota[i]].rson;
             return kth(tree[ra].rson, tree[rb].rson, mid + , r, k - sum);
         }
     }
 }
 //Main operation
 void modify(int k, int val) {
     int x = hash(a[k]), y = hash(val);
     a[k] = val;
     while(k <= n) {
         insert(root2[k], , total - , x, -);
         insert(root2[k], , total - , y, );
         k += lowbit(k);
     }
 }

 int kth(int l, int r, int k) {
     cnta = cntb = ;
     for(int i = l - ; i; i -= lowbit(i)) roota[cnta++] = root2[i];
     for(int i = r; i; i -= lowbit(i)) rootb[cntb++] = root2[i];
     return kth(root[l - ], root[r], , total - , k);
 }

 int main() {
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d", &n, &m);
         for(int i = ; i <= n; ++i) scanf("%d", &a[i]);
         for(int i = ; i <= m; ++i) query[i].read();
         buildHash();
         buildTree();
         initBIT();
         for(int i = ; i <= m; ++i) {
             if(query[i].op == 'Q') printf("%d\n", arr[kth(query[i].i, query[i].j, query[i].k)]);
             else modify(query[i].i, query[i].k);
         }
     }
 }