poj 3368 Frequent values //RMQ

最新推荐文章于 2019-11-21 22:18:02 发布

HQD因为有趣所以做题

最新推荐文章于 2019-11-21 22:18:02 发布

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分类专栏： ACM_数据结构文章标签： each query integer input output

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本文介绍了一种解决区间内最频繁出现数值的查询问题的高效算法。通过离散化和预处理技术，实现了快速查询指定范围内最常出现的整数及其出现次数的功能。

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Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7310		Accepted: 2617

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 10
2 3
1 10
5 10
0

Sample Output

1
4
3

Source

Ulm Local 2007

#include<cstdio>
#include<cstring>
#include<cmath>
#define mmax(a,b) ((a)>(b)?(a):(b))
const int MAXN=100000+10;
int a[MAXN],hash[MAXN];
int f1[MAXN][20];
int t,n;
struct Point
{
    int left,right;
    int count,num;
} point[MAXN];

void init()
//离散化，相同的点缩成一点，记录区间左端点，右端点，出现次数，和数字
{
    point[0].num=a[0];
    point[0].left=0;
    point[0].right=0;
    point[0].count=1;
    hash[0]=0;
    t=0;
    for(int i=2; i<=n; i++)
    {
        if(a[i]==a[i-1])
        {
            point[t].count++;
            point[t].right=i;
        }
        else
        {
            point[++t].left=i;
            point[t].right=i;
            point[t].num=a[i];
            point[t].count=1;
        }
        hash[i]=t;
    }
}
void st(int n)
{
    int i, j, k, m;
    k = (int) (log((double)n) / log(2.0));
    for(i = 0; i < n; i++)
    {
        f1[i][0] = point[i].count; //递推的初值
    }
    for(j = 1; j <= k; j++)
    { //自底向上递推
        for(i = 0; i + (1 << j) - 1 < n; i++)
        {
            m = i + (1 << (j - 1)); //求出中间的那个值
            f1[i][j] = mmax(f1[i][j-1], f1[m][j-1]);
        }
    }
}
//查询i和j之间的最值,注意i是从0开始的
int rmq(int i, int j)
{
    int k = (int)(log(double(j-i+1)) / log(2.0)), t1; //用对2去对数的方法求出k
    return t1 = mmax(f1[i][k], f1[j - (1<<k) + 1][k]);
}
int solve(int a,int b)
{
    if(hash[a]==hash[b])  return b-a+1;
    else if(hash[a]+1==hash[b])
    {
        int l=point[hash[a]].right-a+1;
        int r=b-point[hash[b]].left+1;
        return mmax(l,r);
    }
    else
    {
        int l=point[hash[a]].right-a+1;
        int r=b-point[hash[b]].left+1;
        l=mmax(l,r);
        r=rmq(hash[a]+1,hash[b]-1);
        l=mmax(l,r);
        return l;
    }
}
int main()
{
    int q;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)  break;
        scanf("%d",&q);
        for(int i=0; i<n; ++i)  scanf("%d",&a[i]);
        init();
        st(t);
        while(q--)
        {
            int a,b;
            scanf("%d%d",&a,&b);//在原序列中的区间，判断再转化为线段树上的区间
            printf("%d/n",solve(a-1,b-1));
        }
    }
    return 0;
}