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=100010;
int mm[maxn],Max[maxn][30];
int MAXL[maxn],MAXR[maxn],cnt=0;
int ID[maxn],tot[maxn],a[maxn];
void Init(){
    memset(tot,0,sizeof(tot));
    cnt=0;
}
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=1;i<=n;i++)
            scanf("%d",&a[i]);
        a[0]=a[n+1]=-212429368;
        for(int i=1;i<=n;i++){
            if(a[i]==a[i-1]){    
                MAXL[i]=MAXL[i-1];
                ID[i]=ID[i-1];
            }
            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>=1;i--){
            if(a[i]==a[i+1])
                MAXR[i]=MAXR[i+1];
            else 
                MAXR[i]=i;
        }
        //MAXR[i]指与a[i]相等的i+1~n中最靠右的位置 
        mm[0]=-1;
        for(int i=1;i<=cnt;i++){
            mm[i]=(i&(i-1))?mm[i-1]:mm[i-1]+1;
            Max[i][0]=tot[i];
        } 
        for(int k=1;k<=mm[cnt];k++)
            for(int i=1;i+(1<<k)-1<=cnt;i++)
                Max[i][k]=max(Max[i][k-1],Max[i+(1<<(k-1))][k-1]);
        //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
",b-a+1);
                continue;
            }
            ans=-214748368;
            ans=max(MAXR[a]-a+1,b-MAXL[b]+1);
            a=ID[MAXR[a]+1];
            b=ID[MAXL[b]-1];
            if(a<=b)
                ans=max(ans,max(Max[a][mm[b-a+1]],Max[b-(1<<mm[b-a+1])+1][mm[b-a+1]]));
            printf("%d
",ans);
        }
    }
    return 0;
}