POJ 3368 Frequent values

B - Frequent values

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3368

Appoint description: bjtu_lyc  (2011-08-08)System Crawler  (2015-08-08)

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 andj (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 nintegers 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 10 102 31 105 100

Sample Output

143

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn 100005int n,q;int a[maxn],dp[maxn][30];int value[maxn],cnt[maxn],l[maxn],r[maxn],has[maxn],ll[maxn],rr[maxn];void init_rmq(int n){    for(int i=1;i<n;i++)dp[i][0]=cnt[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<(j-1))-1<n;i++)        dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int rmq(int l,int r){    int k=0;    while((1<<(k+1))<=r-l+1)k++;    return max(dp[l][k],dp[r-(1<<k)+1][k]);}int main(){    int u,v;    //freopen("in.txt","r",stdin);    while(~scanf("%d",&n)&&n){        scanf("%d",&q);        scanf("%d",&a[1]);        int id=1;        memset(value,0,sizeof value);        memset(cnt,0,sizeof cnt);        value[id]=a[1];        cnt[id]++;        has[1]=id;        l[id]=1;        r[id]=1;        for(int i=2;i<=n;i++){            scanf("%d",&a[i]);            if(a[i]==a[i-1]){                    cnt[id]++;            }            else {                id++;                l[id]=i;                value[id]=a[i];                cnt[id]++;            }             has[i]=id;             r[id]=i;        }        for(int i=1;i<=n;i++){            ll[i]=l[has[i]];            rr[i]=r[has[i]];        }        /*        for(int i=1;i<=n;i++){            cout<<has[i]<<" ";        }        cout<<endl;        */        init_rmq(id);        for(int i=0;i<q;i++){            scanf("%d%d",&u,&v);            if(has[u]==has[v])printf("%d\n",v-u+1);            else {                int max1=rr[u]-u+1;            //若不设置ll[]的话，则直接用l[has[u]]即可                int max2=v-ll[v]+1;                int max3=0;                if(has[v]-has[u]>1){                    max3=rmq(has[u]+1,has[v]-1);                }                printf("%d\n",max(max(max1,max2),max3));            }        }    }}