[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[100003],d[100003][50],Left[100003],Right[100003];

void pre()
{
    clr(d,0);
    clr(Left,0);
    clr(Right,0);
    clr(a,0);
}

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

int RMQ_init()
{
    int i,j;
     
    for(j=1;(1<<j)<=n;j++)
        for(i=1;i+(1<<j)-1<=n;i++)
        d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
    
    return 0;
        
}

int RMQ(int L,int R)
{
    int k;
    k=0;
    
    while(1<<(k+1)<=R-L+1) k++;
 
    maxn=max(d[L][k],d[R-(1<<k)+1][k]);;
   
    return 0;
    
}

int Num_init()
{
    int i;

    a[0]=35113;
    rep(i,1,n)
        if(a[i]!=a[i-1]) Left[i]=i;
        else Left[i]=Left[i-1];
  
    a[n+1]=35113;
    red(i,n,1)
        if(a[i]!=a[i+1]) Right[i]=i;
        else Right[i]=Right[i+1];
    
    
    return 0;
}

int main()
{
    int i,x,y;
    
     while(true) {
         pre();    
         scanf("%d",&n); 
        
         if(n==0) exit(0);
         scanf("%d",&q);
         a[0]=35113;
         rep(i,1,n) {
            scanf("%d",&a[i]);
            if(a[i]!=a[i-1]) d[i][0]=1;
            else d[i][0]=d[i-1][0]+1;
            
         }
           
        Num_init();  
        RMQ_init();

        while(q--) {
            scanf("%d%d",&x,&y);
            if(Left[x]==Left[y]) printf("%d\n",y-x+1);
            else { 
                if(Right[x]+1==Left[y]) printf("%d\n",max(Right[x]-x+1,y-Left[y]+1));
                else {       
                     RMQ(Right[x]+1,Left[y]-1);
                     printf("%d\n",max(max(Right[x]-x+1,y-Left[y]+1),maxn));
                }
            }
        }
    }   
    
    
    return 0;
}