POJ 3368 解题报告 RMQ

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10935 Accepted: 4006

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 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 10 102 31 105 100

Sample Output

143

Source

Ulm Local 2007

题目大意：

给出一个连续不降的整数序列，序列长度为n (n<=1e6)，接着给出q次询问，每次询问序列中从l到r中出现次数最多的数的个数。

解题思路：

利用RMQ我的如下，O(N)统计出每个数最右边和它相等的数的位置，记做rt[i]。

如样例中的      1 1 -1 -1 -1 -1 3 10 10 10

统计的rt数组是2 1 6   6  6   6 7  10 10 10

然后再计算出每个数到其最左边和它相等的数的长度

样例所得的数组c[i]为  1 2 1 2 3 4 1 1 2 3。

然后对于每次询问l r，答案就应该是l到其最右边的数的长度，和l最右边的下一个数到r中的c[i]的最大值。

比如对于询问2 4

a[l]的值为1，到其最右边的是i=2，长度为1，然后rt[l]+1=3到r=4中c最大的是2。所以答案就是2。

其中从rt[l]+1到r中求c[i]最大值可以用RMQ（也可以用线段树）来求。

代码供参考：

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int a[100100];int rt[100100];int dp[200100][30];int log2(int x){int res=0;while(x){x/=2;res++;}return res;}int main(){int n,q;while(~scanf("%d",&n)){memset(dp,0,sizeof(dp));memset(rt,0,sizeof(rt));if(n==0)return 0;scanf("%d",&q);for(int i=1;i<=n;i++)scanf("%d",&a[i]);int now=n;rt[n]=n;for(int i=n-1;i>=1;i--){if(a[i]<a[now])now=i;rt[i]=now;}dp[1][0]=1;for(int i=2;i<=n;i++){if(a[i]==a[i-1])dp[i][0]=dp[i-1][0]+1;else dp[i][0]=1;}int log2n=log2(n)-1;for(int i=1;i<=log2n;i++){for(int j=1;j<=n;j++){dp[j][i]=max(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);}}while(q--){int l,r;scanf("%d%d",&l,&r);if(rt[l]>=r)printf("%d\n",r-l+1);else{int tmp=rt[l]-l+1;int len=r-rt[l];int log2len=log2(len)-1;tmp=max(tmp,max(dp[rt[l]+1][log2len],dp[r-(1<<(log2len))+1][log2len]));printf("%d\n",tmp);}}}}