POJ 3368 Frequent values RMQ ST算法/线段树

                                                     Frequent values

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

题意：

给你一个非递减序列。有m次询问，每次询问区间之间出现最多的数字出现的次数。

题解：

RMQ/线段树

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
using namespace std ;
typedef long long ll;
#define inf 100000
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//******************************************************************
struct ss{
int l,r,num,v;
}a[100001];
int counts[100001];
int dp[100001][63],n,m,p;
void rmq_init()
{
   memset(dp,0,sizeof(dp));
   for(int i=1;i<=p;i++){
    dp[i][0]=counts[i];
   }
   int k=floor(log((double)n+1)/log(2.0));
   for(int j=1;j<=k;j++)
   {
       for(int i=1;i+(1<<j)-1<=p;i++)
       {
          int oo=i+(1<<(j-1));
          dp[i][j]=max(dp[i][j-1],dp[oo][j-1]);
       }
   }
}
int RMQ(int x,int y)
{
    int oo=floor(log((double)y-x+1)/log(2.0));
    return max(dp[x][oo],dp[y-(1<<(oo))+1][oo]);
}
void work()
{
    int l,r;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&l,&r);
        if(a[l].num==a[r].num){
            cout<<r-l+1<<endl;
        }
        else {
            int ans=a[l].r-l+1;
            ans=max(r-a[r].l+1,ans);
            l=a[l].num+1;
            r=a[r].num-1;
           if(l<=r) ans=max(ans,RMQ(l,r));
            cout<<ans<<endl;
        }
    }
}
void init()
{

    int f=0,r=0,l=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].v);
        if(f&&a[i].v!=a[i-1].v)
        {
            counts[p]=r-l+1;
            for(int j=l;j<=r;j++)
                a[j].num=p,a[j].l=l,a[j].r=r;
            r++;
            p++;
            l=r;
        }
        else if(f){
            r++;
        }
        if(f==0)
        {
            l=1;
            r=1,f=1;
            p=1;
        }

    }
    counts[p]=r-l+1;
    for(int j=l;j<=r;j++)
                a[j].num=p,a[j].l=l,a[j].r=r;
   // for(int i=1;i<=n;i++)cout<<counts[i]<<" ";
}
int main()
{

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0)break;
        init();
        rmq_init();
        work();
    }
    return 0;
}

RMQ

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
using namespace std ;
typedef long long ll;
#define inf 100000
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//******************************************************************
struct ss
{
    int l,r,v,num;
}tr[100001*7],a[500001];
int counts[200001];
int n,m,p;
void build(int k,int s,int t)
{
    tr[k].l=s;
    tr[k].r=t;
    if(s==t){
        tr[k].v=counts[s];
        return ;
    }
    int mid=(s+t)>>1;
    build(k<<1,s,mid);
    build(k<<1|1,mid+1,t);
    tr[k].v=max(tr[k<<1].v,tr[k<<1|1].v);
}
int ask(int k,int s,int t)
{
    if(s==tr[k].l&&t==tr[k].r)
    {
        return tr[k].v;
    }
    int mid=(tr[k].l+tr[k].r)>>1;
    if(t<=mid)return ask(k<<1,s,t);
    else if(s>mid)return ask(k<<1|1,s,t);
    else {
        return max(ask(k<<1,s,mid),ask(k<<1|1,mid+1,t));
    }
}
void init()
{
    int f=0,l=0,r=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].v);
        if(f&&a[i].v!=a[i-1].v)
        {
            counts[p]=r-l+1;
            for(int j=l;j<=r;j++)a[j].l=l,a[j].r=r,a[j].num=p;

            r++;
            l=r;
            p++;
        }
        else if(f){
            r++;
        }
        if(!f)
        {
            f=l=1;
            r=p=1;
        }
    }
    counts[p]=r-l+1;
   for(int j=l;j<=r;j++)a[j].l=l,a[j].r=r,a[j].num=p;
   ///for(int i=1;i<=p;i++)cout<<counts[i]<<" ";
}
int main()
{

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        p=0;
        if(n==0)break;
        init();
        build(1,1,p);
        // cout<<p<<endl;
           int x,y;
        for(int i=1;i<=m;i++)
        {

            scanf("%d%d",&x,&y);
            if(a[x].num==a[y].num){
                cout<<y-x+1<<endl;
            }
            else {
                int ans=a[x].r-x+1;
                ans=max(ans,y-a[y].l+1);
                x=a[x].num+1;
                y=a[y].num-1;
                if(x<=y)ans=max(ask(1,x,y),ans);
                cout<<ans<<endl;
            }
        }
    }
    return 0;
}

线段树