poj 3368 Frequent values (RMQ或线段树)

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16959 Accepted: 6125

Description

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 integersa₁ , ... , 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 qlines 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 10 102 31 105 100

Sample Output

143

求一段数字区间出现次数最多的数字的频率，数列单调不减。对数列转化后可以用RMQ或者线段树求解。

RMQ经常用来查询区间最值问题，对于任意区间，RMQ算法可以方便的在O(1)时间查询出最值。

以最大值为例，考虑递推方程：f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);

f[i][j]:表示区间[i,i+2^j-1],j=0,1,2,在区间内即可。

上面式子即表示：大区间[i,i+2^j-1]区间=小区间[i,i+2^(j-1)-1]U[i+2^(j-1),i+2^(j-1)+2^(j-1)-1]，

等式两边明显相等（相同），故成立。

对于该题，查询区间[l,r]，对于区间左端点处和区间外面相等的数字单独处理即可。

RMQ解法：

#include<iostream>#include<algorithm>#include<stdio.h>#include<math.h>#include<string.h>using namespace std;#define LL long long int#define rep(i,a,n) for(int i=a;i<n;++i)#define per(i,a,n) for(int i=n-1;i>=a;--i#define N 100005const int inf=1e9+10;int a[N],num[N],rt[N];int dp[N][20];void rmq(int n){    int i,j;    rep(i,1,n+1) dp[i][0]=num[i];    for(j=1;j<20;++j){        for(i=1;i+(1<<(j-1))<=n;++i){            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);        }    }}int myquery(int l,int r){    if(l>r)        return 0;    int k,t1,t2;    k=log(r-l+1.0)/log(2.0);    t1=dp[l][k];    t2=dp[r-(1<<k)+1][k];    return max(t1,t2);}int main(){    //freopen("in.txt","r",stdin);    int n,q;    while(scanf("%d",&n),n){        scanf("%d",&q);        a[0]=inf;        rep(i,1,n+1){            scanf("%d",&a[i]);            if(i==1){                num[i]=1;            }            else if(a[i]==a[i-1]){                num[i]=num[i-1]+1;            }            else{                num[i]=1;            }        }        per(i,1,n+1){            if(a[i]==a[i+1]){                rt[i]=rt[i+1];            }            else{                rt[i]=i;            }        }        rmq(n);        int l,r,t,tmp,ans;        while(q--){            scanf("%d%d",&l,&r);            if(rt[l-1]==rt[l]) t=min(r,rt[l])+1;            else t=l;            tmp=myquery(t,r);            ans=max(t-l,tmp);            cout<<ans<<endl;        }    }    return 0;}