UVA11235:Frequent values(RMQ)

标签：uva

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 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 , ... , 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 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

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。

val[i]代表第i段的数值

cnt[i]代表第i段连续的长度

num[i]表示第i个位置的数在那一段

lef[i],righ[i]分别表示第i段的左右端点位置

所求的最大值就是

1.从L到L所在的段的结束处的元素个数：righ[L]-L+1

2.从R到R所在的段的开始处的元素个数：R-lef[R]+1

3.中间第num[L]+1段到第num[R]-1段的cnt的最大值(RMQ)

答案就是1,2,3中的最大值

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)

int n,q;
int a[N];
int val[N],cnt[N],num[N],lef[N],righ[N];
int d[N][50],len;

void RMQ_init()
{
    int i,j,k;
    for(i = 1; i<=len; i++)
        d[i][0] = cnt[i];
    for(j = 1; (1<<j)<=len+1; j++)
        for(i = 1; i+(1<<j)-1<=len; i++)
            d[i][j] = max(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}

int RMQ(int l,int r)
{
    int k = 0;
    while((1<<(k+1)) <= r-l+1) k++;
    return max(d[l][k],d[r-(1<<k)+1][k]);
}

int main()
{
    int i,j,k,l,r;
    while(~scanf("%d",&n),n)
    {
        scanf("%d",&q);
        len = 1;
        MEM(cnt,0);
        scanf("%d",&a[1]);
        val[len] = a[1];
        cnt[len] = 1;

        num[1] = len;
        lef[0] = 1;
        for(i = 2; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==a[i-1])
            {
                cnt[len]++;
                num[i] = len;
            }
            else
            {
                righ[len] = i-1;
                len++;
                val[len] = a[i];
                cnt[len] = 1;
                num[i] = len;
                lef[len] = i;
            }
        }
        RMQ_init();
        while(q--)
        {
            scanf("%d%d",&l,&r);
            if(num[l]==num[r])
            {
                printf("%d\n",r-l+1);
            }
            else
            {
                int ans=0;
                if(num[l]+1<=num[r]-1)
                    ans=RMQ(num[l]+1,num[r]-1);
                ans = max(ans,max(righ[num[l]]-l+1,r-lef[num[r]]+1));
                printf("%d\n",ans);
            }
        }
    }

    return 0;
}

UVA11235:Frequent values(RMQ)

标签：uva

原文地址：http://blog.csdn.net/libin56842/article/details/46482803