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）：

  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 }

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