Frequent values POJ - 3368

Posted 2020-10-02 啦啦啦

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
------------------------详解请看http://blog.sina.com.cn/s/blog_6635898a0100kzpx.html

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 #define INF 1e8
 7 #define lson l , m , rt << 1
 8 #define rson m+1,r,  rt << 1 | 1
 9 using namespace std;
10 
11 const int maxn=100050;
12 
13 struct mnode{ int l,r,max; }node[4*maxn];
14 struct mnseg{ int sta,end; }seg[maxn]; 
15 
16 int n,q;
17 int num[maxn],hash[maxn];
18 
19 void Inite(){                                        //将序列处理成点的集合
20     for(int i=1;i<=n;i++) scanf("%d",&num[i]);
21     int k=0,pre=999999;
22     for(int i=1;i<=n;i++){
23         if(num[i]!=pre){
24             pre=num[i];
25             k++;
26             seg[k].sta=i;
27             seg[k].end=i;
28         }
29         else seg[k].end=i;
30         hash[i]=k;
31     } 
32 }
33 
34 void Build(int l,int r,int rt){
35     node[rt].l=l;
36     node[rt].r=r;
37     if(l==r){ node[rt].max=seg[l].end-seg[l].sta+1; return; }
38     int m=(l+r)>>1;
39     Build(lson);
40     Build(rson);
41     node[rt].max=max(node[rt<<1].max,node[(rt<<1)|1].max); 
42 }
43 
44 int Query(int l,int r,int rt){                                //key point！！！！
45     if(node[rt].l==l&&node[rt].r==r) return node[rt].max;
46     if(r<=node[rt<<1].r) return Query(l,r,rt<<1);
47     if(l>=node[(rt<<1)|1].l) return Query(l,r,(rt<<1)|1);
48     int a=Query(l,node[rt<<1].r,rt<<1);
49     int b=Query(node[(rt<<1)|1].l,r,(rt<<1)|1);
50     return max(a,b);
51 }
52 
53 int main()
54 {   while(~scanf("%d",&n)){
55         if(n==0) break;
56         scanf("%d",&q);
57         
58         Inite();
59         Build(1,n,1);
60         while(q--){
61             int a,b,pos1,pos2;
62             scanf("%d%d",&a,&b);
63             pos1=hash[a],pos2=hash[b];
64             if(pos1==pos2) cout<<b-a+1<<endl;
65             else{
66                 int n1,n2,n3;
67                 n1=seg[pos1].end-a+1;
68                 n2=0;
69                 n3=b-seg[pos2].sta+1;
70                 if(pos2-pos1>1) n2=Query(pos1+1,pos2-1,1);
71                 cout<<max(n1,max(n2,n3))<<endl;
72             }
73         }
74     }
75     return 0;
76 }