【ZOJ2112】【整体二分+树状数组】带修改区间第k大

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

【分析】

裸题，不说了。

按照这种方法的话，离线的带插入修改区间第K大也应该可以做了。

不过这题的经典作法是树状数组上套可持久化线段树，不过这样空间消耗会很大。

可能要用动态开点？

转一个用块状链表的：http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/19/3268162.html

 /*
 宋代晏殊
 《蝶恋花·槛菊愁烟兰泣露》
 
 槛菊愁烟兰泣露。罗幕轻寒，燕子双飞去。明月不谙离恨苦。斜光到晓穿朱户。
 昨夜西风凋碧树。独上高楼，望尽天涯路。欲寄彩笺兼尺素。山长水阔知何处。
 */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <assert.h>
 #include <map>
 #include <ctime>
 #include <cstdlib>
 #include <stack>
 #define LOCAL
 const int INF = ;
 const int MAXN =  + ;
 using namespace std;
 struct QUERY{
        int x, y;
        int k, s, type, cur;//cur用来记录前面的值
 }q[MAXN], q1[MAXN], q2[MAXN];
 int Ans[MAXN];
 int tmp[MAXN], c[MAXN];
 int n, m, num, cnt;
 int data[MAXN];
 
 inline int lowbit(int x){return x&-x;}
 void add(int x, int val){
      while (x <= n){
            c[x] += val;
            x += lowbit(x);
      }
      return;
 }
 int sum(int x){
     int cnt = ;
     while (x > ){
           cnt += c[x];
           x -= lowbit(x);
     }
     return cnt;
 }
 //整体二分
 void solve(int l, int r, int L, int R){
      //这两个都是结束条件
      if (l > r) return;
      if (L == R){//更新答案
         for (int i = l; i <= r; i++)
         if (q[i].type == ) Ans[q[i].s] = L;
         return;
      }
      int mid = (L + R) >> ;
      for (int i = l; i <= r; i++){
          if (q[i].type ==  && q[i].y <= mid) add(q[i].x, );
          else if (q[i].type ==  && q[i].y <= mid) add(q[i].x, -);
          else if (q[i].type == ) tmp[i] = sum(q[i].y) - sum(q[i].x - );
      }
      //更新完了就要清除标记了
      for (int i = l; i <= r; i++){
          if (q[i].type ==  && q[i].y <= mid) add(q[i].x, -);
          else if (q[i].type ==  && q[i].y <= mid) add(q[i].x, );
      }
      int l1 = , l2 = ;
      for (int i = l; i <= r; i++){
          if (q[i].type == ){
             //不用id就直接改
             if (q[i].cur + tmp[i] > q[i].k - ) q1[++l1] = q[i];
             else {
                  q[i].cur += tmp[i];
                  q2[++l2] = q[i];
             }
          }else{
             if (q[i].y <= mid) q1[++l1] = q[i];
             else q2[++l2] = q[i];
          }
      }
      for (int i = ; i <= l1; i++) q[i + l - ] = q1[i];
      for (int i = ; i <= l2; i++) q[i + l1 + l - ] = q2[i];
      solve(l, l + l1 - , L, mid);
      solve(l + l1, r, mid + , R);
 }
 void init(){
      memset(c, , sizeof(c));
      cnt = num = ;//指针初始化,num记录总的操作数量
      scanf("%d%d", &n, &m);
      for (int i = ; i <= n; i++){
          num++;
          scanf("%d", &data[i]);
          q[num].x = i;q[num].type = ;//1代表插入
          q[num].s = ;q[num].y = data[i];//没有用y就当val用
      }
      for (int i = ; i <= m; i++){
          char str[];
          num++;
          scanf("%s", str);
          if (str[] == 'Q'){
             int l, r, k;
             scanf("%d%d%d", &l, &r, &k);
             q[num].x = l;q[num].y = r;
             q[num].type = ; q[num].s = ++cnt;
             q[num].k = k;
          }else{
             int l, x;
             scanf("%d%d", &l, &x);
             q[num].x = l;q[num].y = data[l];//2为删除
             q[num].type = ;q[num].s = ;
             q[++num].x = l;
             q[num].y = x;//删除后插入
             q[num].type = ;
             q[num].s = ;
             data[l] = x;//注意这里一定要改,不然会影响到后面的更新
          }
      }
      for (int i = ; i <= num; i++) q[i].cur = ;
 }
 
 int main(){
     int T;
 
     scanf("%d", &T);
     while (T--){
           init();
           solve(, num, , INF);
           for (int i = ; i <= cnt; i++) printf("%d\n", Ans[i]);
     }
     return ;
 }