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 = 300010;
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 = 0) {
    while(i > 0) {
        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)>>1,a = 0,b = 0;
    for(int i = lt; i <= rt; ++i) {
        if(Q[i].id) {
            int tmp = sum(Q[i].y) - sum(Q[i].x-1);
            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 = 0; i < a; ++i) Q[lt + i] = A[i];
    for(int i = 0; i < b; ++i) Q[lt + a + i] = B[i];
    solve(lt,lt + a - 1,L,mid);
    solve(lt + a,rt,mid + 1,R);
}
int main() {
    int n,m,cnt,x,y;
    while(~scanf("%d",&n)) {
        memset(c,0,sizeof c);
        cnt = tot = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d",a + i);
            Q[++tot].y = a[i];
            Q[tot].x = i;
            Q[tot].id = 0;
            Q[tot].d = 1;
        }
        scanf("%d",&m);
        for(int i = 1,op; i <= m; ++i) {
            scanf("%d",&op);
            if(op == 2) {
                tot++;
                scanf("%d%d%d",&Q[tot].x,&Q[tot].y,&Q[tot].k);
                Q[tot].id = ++cnt;
                Q[tot].d = 0;
            } else {
                scanf("%d%d",&x,&y);
                Q[++tot].x = x;
                Q[tot].y = a[x];
                Q[tot].id = 0;
                Q[tot].d = -1;

                Q[++tot].x = x;
                Q[tot].y = a[x] = y;
                Q[tot].id = 0;
                Q[tot].d = 1;
            }
        }
        solve(1,tot,0,INF);
        for(int i = 1; i <= cnt; ++i)
            printf("%d
",ans[i]);
    }
    return 0;
}