nbut线段树专题W - 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 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.

这题给我的感受很深，题目给的是一些数，但要求的是出现最多的数的个数（一段区间内），所以如果按数值去建立线段树，显然没法得到答案，所以本题我们必须把几个相同的节点当做一个点，然后建立线段树，由于输入数据为不降序，所以很好统计每一个数的个数，以此建立线段树，注意，线段树只能求出完整的一段区间，所以还要配合情况的判断，另外要注意的是，输入的区间和交给线段树的区间不一样，所以要特殊处理（离散化），我们用结构体数组保存一个数集的起点和终点，再用个hash数组保存每个点属于的数集序号，这样，我们通过点，得到其所属数集，再通过判断是不是属于一个数集，2个或者多个来进行求解最大值（只有多个数集的时候才用到线段树）。

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define maxn 100000
int hash[maxn];
int m,n;
struct node2
{
 int l,r,c;
}tree[maxn<<2];
struct node
{
 int beg,end;
}num[maxn];

void build(int p,int l,int r)
{
 tree[p].l=l;
 tree[p].r=r;
 if(l==r)
 {
 tree[p].c=num[l].end-num[l].beg+1;
 return ;
 }
 int mid=(l+r)>>1;
 build(p<<1,l,mid);
 build(p<<1|1,mid+1,r);
 tree[p].c=max(tree[p<<1].c,tree[p<<1|1].c);
}

int query(int p,int l,int r)
{
 if(l==tree[p].l && r==tree[p].r)
 return tree[p].c;
 int mid=(tree[p].l+tree[p].r)>>1;
 if(r<=mid)
 return query(p<<1,l,r);
 else if(l>mid)
 return query(p<<1|1,l,r);
 else
 return max(query(p<<1,l,mid),query(p<<1|1,mid+1,r));
}

int main()
{
 int n,q,temp;
 while(~scanf("%d",&n))
 {
 if(!n)
 break;
 scanf("%d",&q);
 scanf("%d",&temp);
 memset(num,0,sizeof(num));
 num[1].beg=1;
 num[1].end=1;
 hash[1]=1;
 int k=1,x,y,last=temp;
 for (int i = 2; i <=n; ++i)
 {
 scanf("%d",&temp);
 if(last==temp)
 num[k].end=i;
 else
 {
 k++;
 last=temp;
 num[k].beg=i;
 num[k].end=i;
 }
 hash[i]=k;
 }
 build(1,1,k);
 while(q--)
 {
 scanf("%d%d",&x,&y);
 int s=hash[x],t=hash[y];
 if(s==t)
 printf("%d\n",y-x+1);
 else
 {
 int n1=num[s].end-x+1;
 int n2=0;
 int n3=y-num[t].beg+1;
 if(t-s>1)
 n2=query(1,s+1,t-1);
 printf("%d\n",max(max(n1,n3),n2) );
 }
 }
 }
 return 0;
}