2015 Multi-University Training Contest 10 hdu 5412 CRB and Queries
CRB and Queries
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 533 Accepted Submission(s): 125
Problem Description
There are N boys in CodeLand.
Boy i has his coding skill Ai.
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v.
Query 2: 2 l r k
This is a report query which asks the k-th smallest value of coding skill between Boy l and Boy r(both inclusive).
Input
There are multiple test cases.
The first line contains a single integer N.
Next line contains N space separated integers A1, A2, …, AN, where Ai denotes initial coding skill of Boy i.
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N, Q ≤ 105
1 ≤ Ai, v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1
Output
For each query of type 2, output a single integer corresponding to the answer in a single line.
Sample Input
5
1 2 3 4 5
3
2 2 4 2
1 3 6
2 2 4 2
Sample Output
3
4
Author
KUT(DPRK)
解题:带修改的区间K大查询
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 300010; 4 const int INF = 0x3f3f3f3f; 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 n,m,cnt,x,y; 52 while(~scanf("%d",&n)) { 53 memset(c,0,sizeof c); 54 cnt = tot = 0; 55 for(int i = 1; i <= n; ++i) { 56 scanf("%d",a + i); 57 Q[++tot].y = a[i]; 58 Q[tot].x = i; 59 Q[tot].id = 0; 60 Q[tot].d = 1; 61 } 62 scanf("%d",&m); 63 for(int i = 1,op; i <= m; ++i) { 64 scanf("%d",&op); 65 if(op == 2) { 66 tot++; 67 scanf("%d%d%d",&Q[tot].x,&Q[tot].y,&Q[tot].k); 68 Q[tot].id = ++cnt; 69 Q[tot].d = 0; 70 } else { 71 scanf("%d%d",&x,&y); 72 Q[++tot].x = x; 73 Q[tot].y = a[x]; 74 Q[tot].id = 0; 75 Q[tot].d = -1; 76 77 Q[++tot].x = x; 78 Q[tot].y = a[x] = y; 79 Q[tot].id = 0; 80 Q[tot].d = 1; 81 } 82 } 83 solve(1,tot,0,INF); 84 for(int i = 1; i <= cnt; ++i) 85 printf("%d\n",ans[i]); 86 } 87 return 0; 88 }