UVA - 11235 Frequent values

 

2007/2008 ACM International Collegiate Programming Contest 
University of Ulm Local Contest

 

Problem F: Frequent values

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 Specification

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 Specification

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

---

A naive algorithm may not run in time!

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 const int maxn=100100;
 9 int cnt=0;
10 struct node
11 {
12     int L,R,nb,v;
13 }N[maxn];
14 
15 int dp[maxn+10][20];
16 int a[maxn+10],kt[maxn+10],n,m;
17 
18 int RMQ_init()
19 {
20     for(int i=1;i<=cnt;i++) dp[i][0]=N[i].nb;
21     for(int j=1;(1<<j)<=cnt;j++)
22     {
23         for(int i=1;i+(1<<j)-1<=cnt;i++)
24         {
25             dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
26         }
27     }
28 }
29 
30 int RMQ(int L,int R)
31 {
32     int k=0;
33     while(R-L+1>=(1<<k)) k++; k--;
34     return max(dp[L][k],dp[R-(1<<k)+1][k]);
35 }
36 
37 
38 
39 int main()
40 {
41     while(scanf("%d",&n)&&n)
42     {
43         scanf("%d",&m);
44         ///yu chu li
45         memset(N,0,sizeof(N));
46         scanf("%d",a+1);
47         cnt=0;N[cnt].v=a[1];N[cnt].L=1;N[cnt].nb=1;
48         for(int i=2;i<=n;i++)
49         {
50             scanf("%d",a+i);
51             if(a[i]==N[cnt].v)
52             {
53                 N[cnt].nb++;
54             }
55             else
56             {
57                 cnt++;
58                 N[cnt-1].R=i-1;
59                 N[cnt].L=i;N[cnt].v=a[i];N[cnt].nb=1;
60             }
61             if(i==n) N[cnt].R=n;
62         }
63         int pos=1;
64         kt[0]=-1;
65         for(int i=0;i<=cnt;i++)
66         {
67             int temp=N[i].nb;
68             while(temp--)
69             {
70                 kt[pos]=i;
71                 pos++;
72             }
73         }
74         RMQ_init();
75         while(m--)
76         {
77             int a,b,na=-1,nb=-1;
78             scanf("%d%d",&a,&b);
79             na=kt[a],nb=kt[b];
80             if(na==nb)
81             {
82                 printf("%d\n",b-a+1);
83             }
84             else if(na+1==nb)
85             {
86                 printf("%d\n",max(N[na].R-a+1,b-N[nb].L+1));
87             }
88             else
89             {
90                 int ans1=N[na].R-a+1,ans2=b-N[nb].L+1,ans3=0;
91                 int L=na+1,R=nb-1;
92                 ans3=RMQ(L,R);
93                 printf("%d\n",max(ans1,max(ans2,ans3)));
94             }
95         }
96     }
97     return 0;
98 }