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:

int FrequencyOfX = 0;
for ( int i = 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:

 1 /*

 2 * this code is made by Stomach_ache

 3 * Problem: 1108

 4 * Verdict: Accepted

 5 * Submission Date: 2014-09-04 21:32:52

 6 * Time: 1320MS

 7 * Memory: 4248KB

 8 */

 9 #include <stdio.h>

10 #include <string.h>

11 #include <algorithm>

12 using namespace std;

13 /*Let's fight!!!*/

14   

15 const int Sqrt = 333;

16 const int MAX_N = 101000;

17 int a[MAX_N], ans[MAX_N], freq[MAX_N], cnt[MAX_N];

18 int ll[MAX_N], rr[MAX_N], kk[MAX_N], idx[MAX_N], n, m;

19   

20 bool cmp (int a, int b) {

21     if (ll[a]/Sqrt == ll[b]/Sqrt) return rr[a] < rr[b];

22     return ll[a] < ll[b];

23 }

24   

25 int query(int k) {

26     int lb = 1, ub = 100001;

27     while (ub - lb > 1) {

28         int mid = (lb + ub) / 2;

29         if (freq[mid] >= k) lb = mid;

30         else ub = mid;

31     }

32     return lb;

33 }

34   

35 int main() {

36     int T;

37     scanf("%d", &T);

38     while (T--) {

39         scanf("%d%d", &n, &m);

40         for (int i = 0; i < n; i++) scanf("%d", a+i);

41         for (int i = 0; i < m; i++) {

42             scanf("%d%d%d", ll+i, rr+i, kk+i);

43             idx[i] = i; ll[i]--; rr[i]--;

44         }

45   

46         sort(idx, idx + m, cmp);

47         memset(freq, 0, sizeof(freq));

48         memset(cnt, 0, sizeof(cnt));

49   

50         int cl = 0, cr = -1;

51         for (int i = 0; i < m; i++) {

52             int l = ll[idx[i]], r = rr[idx[i]], k = kk[idx[i]];

53             while (cr < r) { freq[++cnt[a[++cr]]] ++; }

54             while (l < cl) { freq[++cnt[a[--cl]]] ++; }

55             while (r < cr) { freq[cnt[a[cr--]]--] --; }

56             while (cl < l) { freq[cnt[a[cl++]]--] --; }

57             ans[idx[i]] = query(k);

58         }

59   

60         for (int i = 0; i < m; i++) printf("%d\n", ans[i]);

61     }

62   

63     return 0;

64 }