ACdream 1108（莫队）

题目链接

The kth number

Time Limit: 12000/6000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

Do you still remember the Daming Lake's  k'th number? Let me take you back and recall that wonderful memory.

Given a sequence A with length of n,and m querys.Every query is defined by three integer(l,r,k).For each query,please find the kth biggest frequency in interval [l,r].
Frequency of a number x in [l,r] can be defined by this code:

1 2 3 4 5 6

intFrequencyOfX = 0; for(inti = l; i <= r; i ++) {      if(a[i]==X) {          FrequencyOfX ++;      } }

Input

First line is a integer T,the test cases.
For each case:
First line contains two integers n and m.
Second line contains n integers a1,a2,a3....an.
Then next m lines,each line contain three integers l,r,k.

T<=12
1<=n,m,ai<=100000
1<=l<=r<=n
1<=k
data promise that for each query(l,r,k),the kind of number in interval [l,r] is at least k.

Output

for every query,output a integer in a line.

Sample Input

1
6 3
13 14 15 13 14 13
1 6 3
1 6 1
3 5 2

Sample Output

1
3
1

Source

zhangmingming

Manager

wuyiqi

初次接触莫队算法。屠了一次版。。。哈哈，不要在意这些细节

和平方分割的方法类似，莫队算法的思想大概也是把线性的序列尽量平均的进行分割。

一般用于不需要队数据进行修改的题目，而且必须离线。

这里的排序，都是先按照桶的顺序升序排序，如果桶的顺序相同再按终点排序。

Accepted Code:

 /*
 * this code is made by Stomach_ache
 * Problem: 1108
 * Verdict: Accepted
 * Submission Date: 2014-09-04 21:32:52
 * Time: 1320MS
 * Memory: 4248KB
 */
 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 const int Sqrt = ;
 const int MAX_N = ;
 int a[MAX_N], ans[MAX_N], freq[MAX_N], cnt[MAX_N];
 int ll[MAX_N], rr[MAX_N], kk[MAX_N], idx[MAX_N], n, m;

 bool cmp (int a, int b) {
     if (ll[a]/Sqrt == ll[b]/Sqrt) return rr[a] < rr[b];
     return ll[a] < ll[b];
 }

 int query(int k) {
     int lb = , ub = ;
     while (ub - lb > ) {
         int mid = (lb + ub) / ;
         if (freq[mid] >= k) lb = mid;
         else ub = mid;
     }
     return lb;
 }

 int main() {
     int T;
     scanf("%d", &T);
     while (T--) {
         scanf("%d%d", &n, &m);
         for (int i = ; i < n; i++) scanf("%d", a+i);
         for (int i = ; i < m; i++) {
             scanf("%d%d%d", ll+i, rr+i, kk+i);
             idx[i] = i; ll[i]--; rr[i]--;
         }

         sort(idx, idx + m, cmp);
         memset(freq, , sizeof(freq));
         memset(cnt, , sizeof(cnt));

         int cl = , cr = -;
         for (int i = ; i < m; i++) {
             int l = ll[idx[i]], r = rr[idx[i]], k = kk[idx[i]];
             while (cr < r) { freq[++cnt[a[++cr]]] ++; }
             while (l < cl) { freq[++cnt[a[--cl]]] ++; }
             while (r < cr) { freq[cnt[a[cr--]]--] --; }
             while (cl < l) { freq[cnt[a[cl++]]--] --; }
             ans[idx[i]] = query(k);
         }

         for (int i = ; i < m; i++) printf("%d\n", ans[i]);
     }

     return ;
 }