POJ 3368 Frequent values (RMQ)
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Description
You are given a sequence of n integers a1 , a2 , … , an 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 ai , … , aj.
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 , … , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, …, n}) separated by spaces. You can assume that for each i ∈ {1, …, n-1}: ai ≤ ai+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
题意
求区间某个数字出现的最大次数。
思路
题目中有一个很重要的性质,即原序列是不下降的,因此相等的数字一定是挨着的。
以样例来说吧: -1 -1 1 1 1 1 3 10 10 10
。
从左到右扫一遍记录每一个元素左边(包含)连续的相同数字有多少个,则处理后的序列为 1 2 1 2 3 4 1 1 2 3
。
根据这个序列 RMQ 建立 ST 表,我们将其记为
另外我们还需要一个序列 1 1 5 5 5 5 6 9 9 9
。
对于每一个查询
- 假如
al=ar ,说明查询区间都为同一数字,此时区间长度r−l+1 即为结果。 - 假如
al=al−1 ,说明区间左边截断了某种数字的一部分,此时的结果为max(tmpl−l+1,max(ST[tmpl+1],ST[r])) - 对于其他情况,结果为
max(ST[l],ST[r])
AC 代码
#include<iostream>#include<cstdio>using namespace std;const int maxn = 2e5+10;int maxx[30][maxn];int a[maxn];int tmp[maxn];int n,q;void rmq(){ for(int i=1; (1<<i)<=n; i++) for(int j=0; j+(1<<i)<=n; j++) maxx[i][j]=max(maxx[i-1][j],maxx[i-1][j+(1<<(i-1))]);}int query(int l,int r){ int len=0; for(int i=0; 1<<(len+1)<=r-l+1; i++,len++); return max(maxx[len][l-1],maxx[len][r-(1<<len)]);}int main(){ ios::sync_with_stdio(false); cin.tie(0); while(cin>>n&&n) { cin>>q; for(int i=0; i<n; i++) cin>>a[i]; for(int i=0; i<n; i++) { if(i==0||a[i]!=a[i-1]) maxx[0][i] = 1; else maxx[0][i] = maxx[0][i-1]+1; } int ind = n-1; for(int i=n-1; i>=0; i--) { if(a[ind]!=a[i]) ind = i; tmp[i] = ind; } rmq(); for(int i=0; i<q; i++) { int l,r,ans = 0; cin>>l>>r; --l,--r; if(a[l]==a[r]) ans = r-l+1; else if(l>0&&a[l]==a[l-1]) ans = max(tmp[l]-l+1,query(tmp[l]+2,r+1)); else ans = query(l+1,r+1); cout<<ans<<endl; } } return 0;}
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