Uva 11235 Frequent values 【RMQ问题】

Uva 11235 Frequent values 【RMQ问题】http://www.bnuoj.com/bnuoj/problem_show.php?pid=19653

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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【题意】给出一个非递减序列的整数数组a1，...，an，对于一系列的询问（i，j），回答 ai ~ aj 中出现次数最多的值所出现的次数。

【分析】把该序列进行游程编码（Run Length Encoding, RLE），比如,-1,1,1,2,2,2,4可编码成(-1,1),(1,2),(2,3),(4,1)，其中(a,b)表示有b个连续的a。用value[i]和cnt[i]分别表示第i段的数值和出现的次数，num[p],Left[p],Right[p]表示位置p所在段的编号和左右端点位置。每次查询（L,R）的结果是以下三部分的最大值：Right[L]-L+1，R-Left[R]+1，RMQ(num[L]+1, num[R]-1)中的cnt最大值；如果L和R在同一数据段中，则为R-L+1。

【代码如下】

//AcceptedGNU C++232ms2255B#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>using namespace std;const int maxn = 100000+10;int cnt[maxn], num[maxn], Left[maxn], Right[maxn];//cnt[]表示第i个数据段的元素个数, num[]表示位置p所在段的编号//left[]、right[]表示位置p所在段的左右端点位置int d[maxn][20];int n, q, ssize;//ssize表示cnt[]的数据段元素个数void RLE()//读数据，进行游程编码{    int x, pre = INT_MAX;//pre记录当前相同的数据元素    int cur = 0;//当前数据段的位置指针    int l;//l 是当前数据段的最左的边界    for(int i=1; i<=n; i++)    {        scanf("%d", &x);        if(x == pre){            cnt[cur]++; num[i] = cur;            Left[i] = l;        }        else{            if(cur > 0){ //设置右端点                for(int j=l; j<=i-1; j++)                    Right[j] = i-1;            }            cur++;            l = i;            num[i] = cur;            Left[i] = l;            cnt[cur] = 1;            pre = x;        }    }    for(int i=l; i<=n; i++)        Right[i] = n;//设置最后一个数据段的右端点位置    ssize = cur;//共有多少个数据段}//元素从1编号到nvoid RMQ_init(){    for(int i=1; i<=ssize; i++) d[i][0] = cnt[i];    for(int j=1; (1<<j) <= ssize; j++)        for(int i=1; i+j-1 <= ssize; i++)            d[i][j] = max(d[i][j-1], d[i + (1<<(j-1))][j-1]);}int RMQ(int L, int R){    if(L>R) return 0;    int k=0;    while((1<<(k+1)) <= R-L+1) k++;    return max(d[L][k], d[R-(1<<k)+1][k]);}int main(){    while(cin>>n && n!=0)    {        scanf("%d", &q);        RLE();        RMQ_init();        while(q--)        {            int L, R;            scanf("%d %d", &L, &R);            if(num[L] == num[R]) //在相同数据段                printf("%d\n", R-L+1);            else            {                int t1 = Right[L]-L+1;                int t2 = R-Left[R]+1;                int t3 = RMQ(num[L]+1, num[R]-1);                printf("%d\n", max(t1, max(t2, t3)));            }        }    }    return 0;}