UVA 11235 经典RMQ问题

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2176

2007/2008 ACM International Collegiate Programming Contest 
University of Ulm Local Contest

Problem F: Frequent values

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 Specification

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 Specification

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

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A naive algorithm may not run in time!

题目大意：
          给出一个非降序的排列的整数数组a1,a2,a3...an,你的任务是对于一系列的询问（i，j），回答ai,ai+1,....aj,中出现次数最多的值所出现的次数。

解题思路：RMQ范围最小值问题（大白书P197）

          应注意到整个数组是非降序的，所有相等的元素都会聚集到一起。这样就可以把整个数组进行游程编码。比如：-1，1,1,2,2，2,4就可以编码成（-1,1），（1,2），（2,3），（4,1），其中（a，b）表示b个连续的a。用value[i]和count[j]分别表示第i段的数值和出现的次数，num[p],left[p],right[p]分别表示位置p所在段的编号和左右端点的位置，则在下图的情况，每次查询（L,R）的结果为以下三个部分的最大值：从L到L所在段的结束处的元素个数（即right[L]-L+1）、从R所在段的开始处到R处的元素个数（即R-left[R]-1）、中间第num[L]+1段到第num[R]-1段的count的最大值。

#include <stdio.h>#include <iostream>#include <string.h>#include <algorithm>using namespace std;const int maxn=110005;const int inf=2e9;int num[maxn],_left[maxn],_right[maxn],cnt[maxn],d[maxn][30];int pre,n,m;int RLE()//游程编码{    int x,l=0,count=-1;    pre=inf;    int size;    for(int i=0; i<n; i++)    {        scanf("%d",&x);        if(x!=pre)        {            for(int j=l; j<i; j++)                _right[j]=i-1;            l=i;            _left[i]=l;            count++;            pre=x;            num[i]=count;            cnt[count]=1;        }        else        {            _left[i]=l;            num[i]=count;            cnt[count]++;        }    }    for(int i=l; i<n; i++)        _right[i]=n-1;    size=count+1;    return size;    /**    printf("\n\n");    for(int i=0;i<n;i++)        printf("%d ",_left[i]);    printf("\n");    for(int i=0;i<n;i++)        printf("%d ",num[i]);    printf("\n");    for(int i=0;i<n;i++)        printf("%d ",cnt[i]);    printf("\n");    for(int i=0;i<n;i++)        printf("%d ",_right[i]);    printf("\n");    **/}void RMQ_init(){    int n=RLE();    for(int i=0;i<n;i++)        d[i][0]=cnt[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<j)-1<n;i++)            d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);}int RMQ_query(int l,int r){    if(l>r)        return 0;    int k=0;    while((1<<(k+1))<=r-l+1)k++;//如果2^(k+1)<=r-l+1,那么k还可以加1    return max(d[l][k],d[r-(1<<k)+1][k]);}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0)            break;        RMQ_init();        while(m--)        {            int l,r;            scanf("%d%d",&l,&r);            l--;            r--;            if(num[l]==num[r])            {                printf("%d\n",r-l+1);                continue;            }            int t1=_right[l]-l+1;            int t2=r-_left[r]+1;            int t3=RMQ_query(num[l]+1,num[r]-1);            printf("%d\n",max(t1,max(t2,t3)));        }    }    return 0;}