线段树（需要维护左半部分与右半部分）-poj3368

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Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 11921 Accepted: 4360

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 10 102 31 105 100

Sample Output

143线段树写法，代码如下：
#include<iostream>#include<set>#include<map>#include<vector>#include<queue>#include<cmath>#include<climits>#include<cstdio>#include<string>#include<cstring>#include<algorithm>typedef long long LL;using namespace std;struct node{    int s,t,lnum,rnum,mnum;} tree[400010];int N,Q;int d[100110];void build(int s,int t,int num){    tree[num].s=s;    tree[num].t=t;    if(s==t)    {        tree[num].lnum=tree[num].rnum=tree[num].mnum=1;        return ;    }    int mid=(tree[num].s+tree[num].t)>>1;    build(s,mid,num<<1);    build(mid+1,t,(num<<1)+1);    int tmp=0;    if(d[tree[num<<1].t]==d[tree[(num<<1)+1].s])        tmp=tree[num<<1].rnum+tree[(num<<1)+1].lnum;    tree[num].mnum=max(max(tree[num<<1].mnum,tree[(num<<1)+1].mnum),tmp);    tree[num].lnum=tree[num<<1].lnum;    if(tree[num<<1].lnum==mid-s+1&&d[tree[num<<1].t]==d[tree[(num<<1)+1].s])        tree[num].lnum+=tree[(num<<1)+1].lnum;    tree[num].rnum=tree[(num<<1)+1].rnum;    if(tree[(num<<1)+1].rnum==t-mid&&d[tree[(num<<1)+1].s]==d[tree[num<<1].t])        tree[num].rnum+=tree[num<<1].rnum;}int find1(int a,int b,int num){    if(a==tree[num].s&&tree[num].t==b)        return tree[num].mnum;    int mid=(tree[num].s+tree[num].t)>>1;    if(a>mid)        return find1(a,b,(num<<1)+1);    else if(b<=mid)        return find1(a,b,num<<1);    else    {        int l=find1(a,mid,num<<1);        int r=find1(mid+1,b,(num<<1)+1);        int a3=0;         if(d[tree[num*2].t]==d[tree[num*2+1].s])            a3=min(tree[num*2].rnum,mid-a+1)+min(tree[num*2+1].lnum,b-mid);        return max(max(l,r),a3);    }}int main(){    //freopen("in.txt","r",stdin);    int a,b;    while(cin>>N)    {        if(!N)            break;        cin>>Q;        for(int i=1; i<=N; i++)            scanf("%d",&d[i]);        build(1,N,1);        for(int i=0; i<Q; i++)        {            scanf("%d%d",&a,&b);            cout<<find1(a,b,1)<<endl;        }    }    return 0;}