UVA 11235 RMQ问题

最新推荐文章于 2019-10-12 00:11:35 发布

Charon_HN

最新推荐文章于 2019-10-12 00:11:35 发布

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分类专栏： RMQ

本文链接：https://blog.youkuaiyun.com/c___c18/article/details/84308525

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本文介绍了一种针对非降序整数数组的区间查询算法，旨在高效解答一系列询问，即找出指定区间内出现频率最高的数值及其出现次数。通过将相等元素组合并使用二元组表示，结合RMQ问题解决思路，实现快速查询。

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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 Specification
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 Specification
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
A naive algorithm may not run in time!

题意：

给出一个非降序的整数数组，你的任务是对于一系列询问，回答区间内出现最多的值的次数

首先要知道什么是RMQ问题

思路：

因为这个数组是非降序的，所以可以把所有相等的元素组合起来用二元组表示，例如-1,1,1,2,2,2,4就可以表示为(-1,1)(1,2)(2,3)(4,1)，其中(a,b)代表有b个连续的a。

首先将其分段来解决，否则不能套用ST的板子

因此这里用cnt[i]代表当前编号i段的长度，L[i]表示编号为i的段的左端点的位置，R[i]表示编号为i的段的右端点的位置，num[i]表示i这个位置的编号

#include<iostream>
#include<set>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define mem(a,b) memset(a,b,sizeof a);
const int N = 1e5+7;
const int INF=0x3f3f3f3f;
const int P=1e5;
int a[N];
int num[N<<1];
int cnt[N<<1];
int L[N<<1];
int R[N<<1];
int dp[N][43];
void Init(){
    mem(dp,0);//mem(val,0);
    mem(num,0);mem(cnt,0);
    mem(L,0);mem(R,0);
}
void RMQ(int n){
    rep(i,1,n) dp[i][0]=cnt[i];
    for(int j=1;(1<<j)<=n+1;j++){
        for(int i=1;i+(1<<j)-1<=n;i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
    }
}
int query(int l,int r){
    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    int n,q;
    while(scanf("%d",&n)!=EOF,n){
            Init();
            scanf("%d",&q);
            scanf("%d",&a[1]);
            int len=1;
            cnt[len]++;
            num[1]=len;
//            L[0]=1;
            rep(i,2,n) {
            scanf("%d",&a[i]);
            if(a[i]==a[i-1]){
                cnt[len]++;
                num[i]=len;
            }else{
            R[len]=i-1;
            len++;
            cnt[len]=1;
            num[i]=len;
            L[len]=i;
            }
            }
        RMQ(len);
        while(q--){int x,y;
        scanf("%d%d",&x,&y);
        if(num[x]==num[y]){//如果只有一种
            printf("%d\n",y-x+1);
        }else{
            int ans=0;
            if(num[x]+1<=num[y]-1)//如果小于两种就不用RMQ了
                ans=query(num[x]+1,num[y]-1);
            ans=max(ans,max(R[num[x]]-x+1,y-L[num[y]]+1));
            printf("%d\n",ans);
        }
        }
    }
    return 0;
}