POJ3368(RMQ，区间频率问题)

Frequent values

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers a_i , ... , a_j.

Input

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a₁ , ... , a_n (-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the
query.

The last test case is followed by a line containing a single 0.

Output

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

Sample Output

1
4
3

Source

Ulm Local 2007

****************************************************分割线********************************************************************************************

RMQ的两个考法1便是正常求区间最小最大，2是求区间频率大小，然而区间的划分时会造成不完全的划分，我们只要分情况解决就好了，这里我们可以引用数组记录每一个数字出现后的首地址结束位置，

并用数组记录这是第几个数，然后单独计算左右和中间，最后比较就好了，其实还是正常RMQ，代码实现不难但容易出错，要小心计算每一个位置。

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 #include<iostream>
 5 using namespace std;
 6 #define N 12220
 7 int re,r[N],l[N];
 8 int mark[N];
 9 ///mark chucundeshicishushishuyudijigeshu tabashuhedijigeshuxiangduiying
10 ///left chucundeshicishushicongnalikaishi
11 ///rigjt chucundeshicishushicongnalijieshu
12 int num[N];///shijishuzuyijingxioashi,xianzai num baocundeshidejigeshuheyoujige
13 int maxsum[N][20];
14 void RMQ(){///zheyangdeRMQdedabufenchucundedaxiaoshicuowude,zhiyoushaoshucaishizhengquede,womenyaozaisouxunshizhaodaozhengquede.zheyangwomenjiukeyiluanpaile
15     for(int i=1;i<=re;i++) maxsum[i][0]=num[i];
16     for(int j=1;(1<<j)-1<=re;j++){
17         for(int i=1;i+(1<<(j-1))-1<=re;i++)
18             maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<(j-1))-1][j-1]);
19     }
20 }
21 int Max(int a,int b,int c){
22     if(a<b) a=b;
23     if(a<c) a=c;
24     return a;
25 }
26 int main()
27 {
28     int n,q,i,j,k,L,R,val;
29     while(scanf("%d",&n)!=EOF&&n){
30         re=1;
31         scanf("%d%d",&q,&k);
32         val=k;
33         num[1]=1;
34         mark[1]=1;
35         l[0]=1;
36         for(i=2;i<=n;i++){
37             scanf("%d",&k);
38             if(k==val){
39                 num[re]++;//zhesjitongyangdeshudedijige
40                 mark[i]=re;//renweidiyiduan
41             }
42             else{
43                 r[re]=i-1;///soudingshangyigeshudeyoubianjie
44                 re++;///jinruxiayigeduan
45                 mark[i]=re;///xiayiduanbianhao
46                 l[re]=i;///xinyiduankaishidedifang
47                 val=k;///biaojixiaygeshufangbianbijiao
48                 num[re]=1;///xinshudediyigeshukaishi
49             }
50         }
51         RMQ();
52         while(q--){
53             scanf("%d%d",&L,&R);
54             if(mark[L]==mark[R]) printf("%d",R-L+1);//rugouzaitongyiqujian
55             else{
56                 int ans=0;
57                 if(mark[L]+1<=mark[R]-1){//如果两段中存在段
58                     //int k=(int)((log(mark[R]-1))/log(2.0));
59                     int k=(int)((log(mark[R]-mark[L]-1))/log(2.0));
60                     //ans=max(maxsum[mark[L]+1][k],maxsum[mark[R]-(1<<k)][k]);
61                     ans=max(maxsum[mark[L]+1][k],maxsum[mark[R]-(1<<k)][k]);
62                 }
63                 ans=Max(r[mark[L]]-L+1,ans,R-l[mark[R]]+1);
64                 //printf("%d-->%d-->%d
",r[mark[L]]-L+1,ans,R-l[mark[R]]+1);
65                 printf("%d
",ans);
66             }
67         }
68     }
69 }

这里好多注释都是用拼音注释的，不过都很好理解。