[POJ] 3368 / [UVA] 11235 - Frequent values [ST算法]

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 10
2 3
1 10
5 10
0

Sample Output

1
4
3

题解：因为序列a1,a2,a3...an是不下降的，所以相同的数字肯定连续集中在一段。维护d[i][0]为相同数字从左到右分别从1到n开始编号，例如样例中的d[i][0]从左到右为 1，2，1，2，3，4，1，1，2，3。这样就转化成了RMQ问题。因为没有中途修改a[i]的值，所以可以用Sparse-Table算法，d[i][j]表示从i开始长度为2^i的一段元素的最大值。维护数组Left[i]，Right[i]分别为相同数字连续一段最左端的编号和最右端的编号，例如样例中的Left[i]分别为：1，1，3，3，3，3，7，8，8，8。所以，序列分布共有以下几种情况：

　　　　　　　　　　　　　　
　　　　　[1]   ans=max(Right[i]-L+1,R-Left[j]+1)
　　　　　[2]   ans=R-L+1

　　　　  [3]   ans=max(max(Right[i]-L+1,R-Left[j]+1),RMQ(k))

代码：

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<ctype.h>
 #include<stdlib.h>
 #include<stdbool.h>

 #define rep(i,a,b)      for(i=(a);i<=(b);i++)
 #define red(i,a,b)      for(i=(a);i>=(b);i--)
 #define clr(x,y)        memset(x,y,sizeof(x))
 #define sqr(x)          (x*x)
 #define LL              long long

 int i,j,n,m,maxn,q,
     a[],d[][],Left[],Right[];

 void pre()
 {
     clr(d,);
     clr(Left,);
     clr(Right,);
     clr(a,);
 }

 int max(int a,int b)
 {
     if(a>b) return a;
     return b;
 }

 int RMQ_init()
 {
     int i,j;

     for(j=;(<<j)<=n;j++)
         for(i=;i+(<<j)-<=n;i++)
         d[i][j]=max(d[i][j-],d[i+(<<(j-))][j-]);

     return ;

 }

 int RMQ(int L,int R)
 {
     int k;
     k=;

     while(<<(k+)<=R-L+) k++;

     maxn=max(d[L][k],d[R-(<<k)+][k]);;

     return ;

 }

 int Num_init()
 {
     int i;

     a[]=;
     rep(i,,n)
         if(a[i]!=a[i-]) Left[i]=i;
         else Left[i]=Left[i-];

     a[n+]=;
     red(i,n,)
         if(a[i]!=a[i+]) Right[i]=i;
         else Right[i]=Right[i+];

     return ;
 }

 int main()
 {
     int i,x,y;

      while(true) {
          pre();
          scanf("%d",&n); 

          if(n==) exit();
          scanf("%d",&q);
          a[]=;
          rep(i,,n) {
             scanf("%d",&a[i]);
             if(a[i]!=a[i-]) d[i][]=;
             else d[i][]=d[i-][]+;

          }

         Num_init();
         RMQ_init();

         while(q--) {
             scanf("%d%d",&x,&y);
             if(Left[x]==Left[y]) printf("%d\n",y-x+);
             else {
                 if(Right[x]+==Left[y]) printf("%d\n",max(Right[x]-x+,y-Left[y]+));
                 else {
                      RMQ(Right[x]+,Left[y]-);
                      printf("%d\n",max(max(Right[x]-x+,y-Left[y]+),maxn));
                 }
             }
         }
     }   

     return ;
 }