本文主要是介绍poj 3368 Frequent values，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：
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Description
You are given a sequence of n integers a1 , a2 , ... , an 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 ai , ... , aj.
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 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+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
题目大意： 给出非下降的序列，q次询问某个区间内数最多的值是多少数（比如在区间1-10内最多的是-1，有4个，则输出4）
思路： 离散化+RMQ,求离散后的最大值在对应到相应的区间上
#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>using namespace std;int M[100000];///存线段树的节点
int A[100000];///存每个值的长度（相当于离散化）
int pre[100000];///存输入区间值所对应的离散后的区间
int l[100000];///每个值的左边界
int r[100000];///右边界
int a[100000];
int n,q;
int len;///长度
///建线段树
void build_RMQ(int node,int b,int e,int M[],int A[])
{if(b==e){M[node]=A[b];}else{build_RMQ(2*node,b,(b+e)/2,M,A);build_RMQ(2*node+1,(b+e)/2+1,e,M,A);if(M[2*node]>=M[2*node+1])M[node]=M[2*node];else M[node]=M[2*node+1];}
}
///查询线段树区间取最大值
int query(int node,int b,int e,int M[],int A[],int i,int j)
{int p1,p2;if(i>e||j<b)return -1;if(b>=i&&e<=j)return M[node];p1=query(2*node,b,(b+e)/2,M,A,i,j);p2=query(2*node+1,(b+e)/2+1,e,M,A,i,j);if(p1==-1)return p2;if(p2==-1)return p1;if(p1<=p2)return p2;return p1;
}
int main()
{while(scanf("%d",&n)){memset(M,0,sizeof(0));if(!n)break;scanf("%d",&q);scanf("%d",&a[1]);memset(A,0,sizeof(A));len=1;A[len]=1;l[len]=1;pre[1]=len;for(int i=2; i<=n; i++){scanf("%d",&a[i]);if(a[i]==a[i-1]){A[len]++;pre[i]=len;}else{r[len]=i-1;len++;l[len]=i;pre[i]=len;A[len]=1;}//cout<<"**"<<endl;}//cout<<"#"<<A[1]<<" "<<A[2]<<endl;//r[len]=n;// cout<<len<<endl;/* for(int i=1;i<=len;i++){cout<<i<<" "<<A[i]<<endl;}*/build_RMQ(1,1,len,M,A);int u,v;while(q--){scanf("%d%d",&u,&v);int c=pre[u];int d=pre[v];int b=pre[u]+1;///把输入的区间去掉（因为不知道输入的///是否包含了该值的所有长度）int e=pre[v]-1;//<<b<<" "<<e<<endl;if(c==d){printf("%d\n",v-u+1);}///特殊情况处理else if(c+1==d){printf("%d\n",max(r[c]-u+1,v-l[d]+1));}else{int mx=query(1,1,len,M,A,b,e);//cout<<"mx:"<<mx<<endl;//cout<<"r[]:"<<r[pre[u]]<<endl;//cout<<"l[]:"<<l[pre[v]]<<endl;int k=r[pre[u]]-u+1;int h=v-l[pre[v]]+1;///在比较中间的最大长度的值和残余的两个边界的长度printf("%d\n",max(mx,max(k,h)));}}}return 0;
}
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