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）

Source

2015 Multi-University Training Contest 10 1007

解题：带修改的区间K大查询

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 const int INF = 0x3f3f3f3f;
 struct QU {
     int x,y,k,id,d;
 } Q[maxn],A[maxn],B[maxn];
 int a[maxn],c[maxn],ans[maxn],tot;
 void add(int i,int val) {
     while(i < maxn) {
         c[i] += val;
         i += i&-i;
     }
 }
 int sum(int i,int ret = ) {
     while(i > ) {
         ret += c[i];
         i -= i&-i;
     }
     return ret;
 }
 void solve(int lt,int rt,int L,int R) {
     if(lt > rt) return;
     if(L == R) {
         for(int i = lt; i <= rt; ++i)
             if(Q[i].id) ans[Q[i].id] = L;
         return;
     }
     int mid = (L + R)>>,a = ,b = ;
     for(int i = lt; i <= rt; ++i) {
         if(Q[i].id) {
             int tmp = sum(Q[i].y) - sum(Q[i].x-);
             if(Q[i].d + tmp >= Q[i].k) A[a++] = Q[i];
             else {
                 Q[i].d += tmp;
                 B[b++] = Q[i];
             }
         } else if(Q[i].y <= mid) {
             add(Q[i].x,Q[i].d);
             A[a++] = Q[i];
         } else B[b++] = Q[i];
     }
     for(int i = lt; i <= rt; ++i)
         if(!Q[i].id && Q[i].y <= mid) add(Q[i].x,-Q[i].d);
     for(int i = ; i < a; ++i) Q[lt + i] = A[i];
     for(int i = ; i < b; ++i) Q[lt + a + i] = B[i];
     solve(lt,lt + a - ,L,mid);
     solve(lt + a,rt,mid + ,R);
 }
 int main() {
     int n,m,cnt,x,y;
     while(~scanf("%d",&n)) {
         memset(c,,sizeof c);
         cnt = tot = ;
         for(int i = ; i <= n; ++i) {
             scanf("%d",a + i);
             Q[++tot].y = a[i];
             Q[tot].x = i;
             Q[tot].id = ;
             Q[tot].d = ;
         }
         scanf("%d",&m);
         for(int i = ,op; i <= m; ++i) {
             scanf("%d",&op);
             if(op == ) {
                 tot++;
                 scanf("%d%d%d",&Q[tot].x,&Q[tot].y,&Q[tot].k);
                 Q[tot].id = ++cnt;
                 Q[tot].d = ;
             } else {
                 scanf("%d%d",&x,&y);
                 Q[++tot].x = x;
                 Q[tot].y = a[x];
                 Q[tot].id = ;
                 Q[tot].d = -;

                 Q[++tot].x = x;
                 Q[tot].y = a[x] = y;
                 Q[tot].id = ;
                 Q[tot].d = ;
             }
         }
         solve(,tot,,INF);
         for(int i = ; i <= cnt; ++i)
             printf("%d\n",ans[i]);
     }
     return ;
 }