数据结构(RMQ)：UVAoj 11235 Frequent values

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

　　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

　　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

　　这题由于序列非降，所以就可以把整个数组进行游程编码(RLE Run Length Encoding)。
　　还是很简单的~~~

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 int mm[maxn],Max[maxn][];
 int MAXL[maxn],MAXR[maxn],cnt=;
 int ID[maxn],tot[maxn],a[maxn];
 void Init(){
     memset(tot,,sizeof(tot));
     cnt=;
 }
 int main(){
 #ifndef ONLINE_JUDGE
     //freopen(".in","r",stdin);
     //freopen(".out","w",stdout);
 #endif

     int n,Q;
     while(~scanf("%d",&n)&&n){
         scanf("%d",&Q);
         Init();
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         a[]=a[n+]=-;
         for(int i=;i<=n;i++){
             if(a[i]==a[i-]){
                 MAXL[i]=MAXL[i-];
                 ID[i]=ID[i-];
             }
             else{
                 MAXL[i]=i;
                 ID[i]=++cnt;
             }
             tot[ID[i]]++;
         }
         //MAXL[i]指与a[i]相等的1~i-1中最靠左的位置
         //ID[i]指当前位置被分到第几块
         //tot[i]指第i块的长度
         for(int i=n;i>=;i--){
             if(a[i]==a[i+])
                 MAXR[i]=MAXR[i+];
             else
                 MAXR[i]=i;
         }
         //MAXR[i]指与a[i]相等的i+1~n中最靠右的位置
         mm[]=-;
         for(int i=;i<=cnt;i++){
             mm[i]=(i&(i-))?mm[i-]:mm[i-]+;
             Max[i][]=tot[i];
         }
         for(int k=;k<=mm[cnt];k++)
             for(int i=;i+(<<k)-<=cnt;i++)
                 Max[i][k]=max(Max[i][k-],Max[i+(<<(k-))][k-]);
         //对块处理出稀疏表
         int a,b,ans;
         while(Q--){
             scanf("%d%d",&a,&b);
             if(a>b)swap(a,b);
             if(ID[a]==ID[b]){
                 printf("%d\n",b-a+);
                 continue;
             }
             ans=-;
             ans=max(MAXR[a]-a+,b-MAXL[b]+);
             a=ID[MAXR[a]+];
             b=ID[MAXL[b]-];
             if(a<=b)
                 ans=max(ans,max(Max[a][mm[b-a+]],Max[b-(<<mm[b-a+])+][mm[b-a+]]));
             printf("%d\n",ans);
         }
     }
     return ;
 }