RMQ ST算法 uva11235（蓝书例题）

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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

A naive algorithm may not run in time!

题意：

给出一个非降序的整数数组，你的任务是对于一系列询问，回答区间内出现最多的值的次数

思路：

因为这个数组是非降序的，所以可以把所有相等的元素组合起来用二元组表示，例如-1,1,1,2,2,2,4就可以表示为(-1,1)(1,2)(2,3)(4,1)，其中(a,b)代表有b个连续的a。

val[i]代表第i段的数值

cnt[i]代表第i段连续的长度

num[i]表示第i个位置的数在那一段

lef[i],righ[i]分别表示第i段的左右端点位置

所求的最大值就是

1.从L到L所在的段的结束处的元素个数：righ[L]-L+1

2.从R到R所在的段的开始处的元素个数：R-lef[R]+1

3.中间第num[L]+1段到第num[R]-1段的cnt的最大值(RMQ)

答案就是1,2,3中的最大值

#include <iostream>  #include <stdio.h>  #include <string.h>  #include <stack>  #include <queue>  #include <map>  #include <set>  #include <vector>  #include <math.h>  #include <bitset>  #include <algorithm>  #include <climits>  using namespace std;    #define LS 2*i  #define RS 2*i+1  #define UP(i,x,y) for(i=x;i<=y;i++)  #define DOWN(i,x,y) for(i=x;i>=y;i--)  #define MEM(a,x) memset(a,x,sizeof(a))  #define W(a) while(a)  #define gcd(a,b) __gcd(a,b)  #define LL long long  #define N 100005  #define MOD 1000000007  #define INF 0x3f3f3f3f  #define EXP 1e-8  #define lowbit(x) (x&-x)    int n,q;  int a[N];  int val[N],cnt[N],num[N],lef[N],righ[N];  int d[N][50],len;    void RMQ_init()  {      int i,j,k;      for(i = 1; i<=len; i++)          d[i][0] = cnt[i];      for(j = 1; (1<<j)<=len+1; j++)          for(i = 1; i+(1<<j)-1<=len; i++)              d[i][j] = max(d[i][j-1],d[i+(1<<(j-1))][j-1]);  }    int RMQ(int l,int r)  {      int k = 0;      while((1<<(k+1)) <= r-l+1) k++;      return max(d[l][k],d[r-(1<<k)+1][k]);  }    int main()  {      int i,j,k,l,r;      while(~scanf("%d",&n),n)      {          scanf("%d",&q);          len = 1;          MEM(cnt,0);          scanf("%d",&a[1]);          val[len] = a[1];          cnt[len] = 1;            num[1] = len;          lef[0] = 1;          for(i = 2; i<=n; i++)          {              scanf("%d",&a[i]);              if(a[i]==a[i-1])              {                  cnt[len]++;                  num[i] = len;              }              else              {                  righ[len] = i-1;                  len++;                  val[len] = a[i];                  cnt[len] = 1;                  num[i] = len;                  lef[len] = i;              }          }          RMQ_init();          while(q--)          {              scanf("%d%d",&l,&r);              if(num[l]==num[r])              {                  printf("%d\n",r-l+1);              }              else              {                  int ans=0;                  if(num[l]+1<=num[r]-1)                      ans=RMQ(num[l]+1,num[r]-1);                  ans = max(ans,max(righ[num[l]]-l+1,r-lef[num[r]]+1));                  printf("%d\n",ans);              }          }      }        return 0;  }

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