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）：
1 #include <cstdio>
2 #include <iostream>
3 #include <algorithm>
4 #include <cstring>
5 using namespace std;
6 typedef long long LL;
7
8 const int MAXN = 50010;
9 const int MAXM = 10010;
10 const int MAXT = MAXM * 15 * 16;
11
12 struct Query {
13 char op;
14 int i, j, k;
15 void read() {
16 scanf(" %c", &op);
17 if(op == ‘Q‘) scanf("%d%d%d", &i, &j, &k);
18 else scanf("%d%d", &i, &k);
19 }
20 } query[MAXM];
21 int a[MAXN];
22 int n, m, T;
23 //hashmap
24 int arr[MAXN + MAXM], total;
25
26 void buildHash() {
27 total = 0;
28 for(int i = 1; i <= n; ++i) arr[total++] = a[i];
29 for(int i = 1; i <= m; ++i)
30 if(query[i].op == ‘C‘) arr[total++] = query[i].k;
31 sort(arr, arr + total);
32 total = unique(arr, arr + total) - arr;
33 }
34
35 int hash(int x) {
36 return lower_bound(arr, arr + total, x) - arr;
37 }
38 //Chairman tree
39 struct Node {
40 int lson, rson, sum;
41 void clear() {
42 lson = rson = sum = 0;
43 }
44 } tree[MAXT];
45 int root[MAXN];
46 int tcnt;
47
48 void insert(int& rt, int l, int r, int pos) {
49 tree[tcnt] = tree[rt]; rt = tcnt++;
50 tree[rt].sum++;
51 if(l < r) {
52 int mid = (l + r) >> 1;
53 if(pos <= mid) insert(tree[rt].lson, l, mid, pos);
54 else insert(tree[rt].rson, mid + 1, r, pos);
55 }
56 }
57
58 void buildTree() {
59 tcnt = 1;
60 for(int i = 1; i <= n; ++i) {
61 root[i] = root[i - 1];
62 insert(root[i], 0, total - 1, hash(a[i]));
63 }
64 }
65 //Binary Indexed Trees
66 int root2[MAXN];
67 int roota[50], rootb[50];
68 int cnta, cntb;
69
70 void initBIT() {
71 for(int i = 1; i <= n; ++i) root2[i] = 0;
72 }
73
74 inline int lowbit(int x) {
75 return x & -x;
76 }
77
78 void insert(int& rt, int l, int r, int pos, int val) {
79 if(rt == 0) tree[rt = tcnt++].clear();
80 if(l == r) {
81 tree[rt].sum += val;
82 } else {
83 int mid = (l + r) >> 1;
84 if(pos <= mid) insert(tree[rt].lson, l, mid, pos, val);
85 else insert(tree[rt].rson, mid + 1, r, pos, val);
86 tree[rt].sum = tree[tree[rt].lson].sum + tree[tree[rt].rson].sum;
87 }
88 }
89
90 int kth(int ra, int rb, int l, int r, int k) {
91 if(l == r) {
92 return l;
93 } else {
94 int sum = tree[tree[rb].lson].sum - tree[tree[ra].lson].sum, mid = (l + r) >> 1;
95 for(int i = 0; i < cntb; ++i) sum += tree[tree[rootb[i]].lson].sum;
96 for(int i = 0; i < cnta; ++i) sum -= tree[tree[roota[i]].lson].sum;
97 if(sum >= k) {
98 for(int i = 0; i < cntb; ++i) rootb[i] = tree[rootb[i]].lson;
99 for(int i = 0; i < cnta; ++i) roota[i] = tree[roota[i]].lson;
100 return kth(tree[ra].lson, tree[rb].lson, l, mid, k);
101 } else {
102 for(int i = 0; i < cntb; ++i) rootb[i] = tree[rootb[i]].rson;
103 for(int i = 0; i < cnta; ++i) roota[i] = tree[roota[i]].rson;
104 return kth(tree[ra].rson, tree[rb].rson, mid + 1, r, k - sum);
105 }
106 }
107 }
108 //Main operation
109 void modify(int k, int val) {
110 int x = hash(a[k]), y = hash(val);
111 a[k] = val;
112 while(k <= n) {
113 insert(root2[k], 0, total - 1, x, -1);
114 insert(root2[k], 0, total - 1, y, 1);
115 k += lowbit(k);
116 }
117 }
118
119 int kth(int l, int r, int k) {
120 cnta = cntb = 0;
121 for(int i = l - 1; i; i -= lowbit(i)) roota[cnta++] = root2[i];
122 for(int i = r; i; i -= lowbit(i)) rootb[cntb++] = root2[i];
123 return kth(root[l - 1], root[r], 0, total - 1, k);
124 }
125
126 int main() {
127 scanf("%d", &T);
128 while(T--) {
129 scanf("%d%d", &n, &m);
130 for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
131 for(int i = 1; i <= m; ++i) query[i].read();
132 buildHash();
133 buildTree();
134 initBIT();
135 for(int i = 1; i <= m; ++i) {
136 if(query[i].op == ‘Q‘) printf("%d\n", arr[kth(query[i].i, query[i].j, query[i].k)]);
137 else modify(query[i].i, query[i].k);
138 }
139 }
140 }
ZOJ 2112 Dynamic Rankings（主席树の动态kth）,布布扣,bubuko.com