Dynamic Rankings
Time Limit: 10000ms
Memory Limit: 32768KB
This problem will be judged on ZJU. Original ID: 2112
64-bit integer IO format: %lld Java class name: Main
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
(adviser)
Site: http://zhuzeyuan.hp.infoseek.co.jp/index.files/our_contest_20040619.htm
Source
Online Contest of Christopher‘s Adventure
解题：整体二分真是叼
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 300010;
4 const int INF = 1000000000;
5 struct QU {
6 int x,y,k,id,d;
7 } Q[maxn],A[maxn],B[maxn];
8 int a[maxn],c[maxn],ans[maxn],tot;
9 void add(int i,int val) {
10 while(i < maxn) {
11 c[i] += val;
12 i += i&-i;
13 }
14 }
15 int sum(int i,int ret = 0) {
16 while(i > 0) {
17 ret += c[i];
18 i -= i&-i;
19 }
20 return ret;
21 }
22 void solve(int lt,int rt,int L,int R) {
23 if(lt > rt) return;
24 if(L == R) {
25 for(int i = lt; i <= rt; ++i)
26 if(Q[i].id) ans[Q[i].id] = L;
27 return;
28 }
29 int mid = (L + R)>>1,a = 0,b = 0;
30 for(int i = lt; i <= rt; ++i) {
31 if(Q[i].id) {
32 int tmp = sum(Q[i].y) - sum(Q[i].x-1);
33 if(Q[i].d + tmp >= Q[i].k) A[a++] = Q[i];
34 else {
35 Q[i].d += tmp;
36 B[b++] = Q[i];
37 }
38 } else if(Q[i].y <= mid) {
39 add(Q[i].x,Q[i].d);
40 A[a++] = Q[i];
41 } else B[b++] = Q[i];
42 }
43 for(int i = lt; i <= rt; ++i)
44 if(!Q[i].id && Q[i].y <= mid) add(Q[i].x,-Q[i].d);
45 for(int i = 0; i < a; ++i) Q[lt + i] = A[i];
46 for(int i = 0; i < b; ++i) Q[lt + a + i] = B[i];
47 solve(lt,lt + a - 1,L,mid);
48 solve(lt + a,rt,mid + 1,R);
49 }
50 int main() {
51 int kase,n,m,cnt,x,y;
52 scanf("%d",&kase);
53 while(kase--) {
54 scanf("%d%d",&n,&m);
55 memset(c,0,sizeof c);
56 cnt = tot = 0;
57 for(int i = 1; i <= n; ++i) {
58 scanf("%d",a + i);
59 Q[++tot].y = a[i];
60 Q[tot].x = i;
61 Q[tot].id = 0;
62 Q[tot].d = 1;
63 }
64 char op[3];
65 for(int i = 1; i <= m; ++i) {
66 scanf("%s",op);
67 if(op[0] == ‘Q‘) {
68 tot++;
69 scanf("%d%d%d",&Q[tot].x,&Q[tot].y,&Q[tot].k);
70 Q[tot].id = ++cnt;
71 Q[tot].d = 0;
72 } else {
73 scanf("%d%d",&x,&y);
74 Q[++tot].x = x;
75 Q[tot].y = a[x];
76 Q[tot].id = 0;
77 Q[tot].d = -1;
78
79 Q[++tot].x = x;
80 Q[tot].y = a[x] = y;
81 Q[tot].id = 0;
82 Q[tot].d = 1;
83 }
84 }
85 solve(1,tot,0,INF);
86 for(int i = 1; i <= cnt; ++i)
87 printf("%d\n",ans[i]);
88 }
89 return 0;
90 }