RMQ（连续相同最大值）

http://poj.org/problem?id=3368

Frequent values

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24727		Accepted: 8612

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

题意：排好序的序列，询问区间值连续出现最多的次数。st表

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
using namespace std;
const int N = 1e5 + 9 ;
const int M = 1e5 + 9 ;
int dp[N][40] , a[N] , b[N] ;//b数组统计连续相同个数
int n , q ;

void RMQ()
{
    for(int i = 1 ; i <= n ; i++)
    {
        dp[i][0] = b[i];
    }
    for(int j = 1 ; j < 30 ; j++)
    {
        for(int i = 1 ; i <= n ; i++)
        {
            if(i + (1 << j) - 1 <= n)
            {
                dp[i][j] = max(dp[i][j-1] , dp[i+(1<<(j-1))][j-1]);
            }
        }
    }

}

int query(int l , int r)
{
    if(r < l) return 0 ;//这个条件不能少，如果查询左右端点相等时，l将大于r
    int ans = 0 ;
    int k = (int)log2(r - l + 1);
    ans = max(dp[l][k] , dp[r - (1 << k) + 1][k]);
    return ans ;
}

int main()
{

    while(~scanf("%d" , &n) && n)
    {

        scanf("%d" , &q);
        //init();
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d" , &a[i]);
            b[i] = 1 ;
        }
        for(int i = 2 ; i <= n ; i++)
        {
            if(a[i] == a[i-1])
            {
                b[i] = b[i-1] + 1 ;
            }
            else
            {
                b[i] = 1 ;
            }
        }
        RMQ();
        for(int i = 1 ; i <= q ; i++)
        {
            int l ,  r , ans = 0;
            scanf("%d%d" , &l , &r);
            int t = l ;
            while(t <= r && a[t] == a[t-1])//以防左端点被截断
            {
                t++;
            }
            ans = max(t - l , query(t , r));//左端点被截断的话，需要用值减去下标
            printf("%d
" , ans);
        }

    }





    return 0 ;
}

线段树

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
using namespace std;
const int N = 1e5 + 9 ;
const int M = 1e5 + 9 ;
int dp[N][40] , a[N] , b[N] ;//b数组统计连续相同个数
int n , q , k = 1 , ans = 0;

struct node{
    int l , r , val ;
}tree[N << 2];

void build(int l , int r , int root)
{
    tree[root].l = l , tree[root].r = r ;
    if(l == r)
    {
        tree[root].val = b[k++];
      //  cout << root << " " << tree[root].val << endl ;
        return ;
    }
    int mid = (l + r) >> 1;
    build(l , mid , root*2);
    build(mid+1 , r , root*2+1);
    tree[root].val = max(tree[root*2].val , tree[root*2+1].val);
}

void query(int l , int r , int root)
{
    if(r < l)
        return  ;
    if(tree[root].l >= l && tree[root].r <= r)
    {
        ans = max(tree[root].val , ans) ;
        return ;
    }
    int mid =(tree[root].l + tree[root].r) >> 1;
    if(l <= mid)
        query(l , r , root*2);
    if(r > mid)
        query(l , r , root*2+1);
}


int main()
{

    while(~scanf("%d" , &n) && n)
    {

        scanf("%d" , &q);
        //init();
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d" , &a[i]);
            b[i] = 1 ;
        }
        for(int i = 2 ; i <= n ; i++)
        {
            if(a[i] == a[i-1])
            {
                b[i] = b[i-1] + 1 ;
            }
            else
            {
                b[i] = 1 ;
            }
        }
      //  memset(tree , 0 ,sizeof(tree));
        k = 1 ;
        build(1 , n , 1);
        for(int i = 1 ; i <= q ; i++)
        {
            int l ,  r , sum = 0;
            scanf("%d%d" , &l , &r);
            int t = l ;
            while(t <= r && a[t] == a[t-1])//以防左端点被截断
            {
                t++;
            }
            query(t , r , 1);

            sum = max(ans , t - l);
            printf("%d
" , sum);
            ans = 0 ;
        }

    }





    return 0 ;
}