UVA 11235 Frequent values

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本文详细解析了如何解决RMQ问题，并首次应用游程编码技术。通过离散化数据，统计下标与编号的关系，实现区间查询效率的提升。特别关注于l与r同端情况下的答案计算，提供了具体的AC代码实例。

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题意：给出长度为n的递增数列，求出m-n间的最大众数的值

链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=23846

思路：第一个RMQ问题，看着小白书上的代码打的。也是第一次遇到游程编码的应用。先对数据进行离散化，分别统计下标p的编号及左端与右端的编号，每次查询通过拆分区间成两个2^i与2^j（因为是求最值，查询可重复），分别查询i区间和j区间，去其中的最大值

注意点：l与r同端答案为r-l+1

以下为AC代码：

RunID	User	OJ All	Prob ID	Result All	Memory (KB)	Time (ms)	Language All	Length (Bytes)	Submit Time
3121775	luminous11	UVA	11235	Accepted	0	1212	C++11 4.8.2	2466	2014-12-29 23:45:35

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;

const int maxn = 100005;
vector<int> cnt;

int m, n, a[maxn];
int d[maxn][18];
int l[maxn];
int r[maxn];
int num[maxn];

void rmq_init ()
{
    int nc = cnt.size();
    for ( int i = 0; i < nc; i ++ )
    {
        d[i+1][0] = cnt[i];
    }
    for ( int j = 1; ( 1 << j ) <= nc; j ++ )
    {
        for ( int i = 1; i + ( 1 << j ) - 1 <= nc; i ++ )
        {
            d[i][j] = max ( d[i][j-1], d[i+(1<<(j-1))][j-1] );
            //cout << "i = " << i << " j = " << j << " d[i][j] = " << d[i][j] << endl;
        }
    }
}

int rmq_init()
{
    int len = cnt.size();
    for ( int i = 0; i < nc; i ++ )
    {
        d[i+1][0] = cnt[i];
    }
    for ( int j = 1; ( 1 << j ) <= nc; j ++ )
    {
        for ( int i = 1; i + ( 1 << j ) - 1 <= nc; i ++ )
            d[i][j] = max ( d[i][j-1], d[i+(1<<(j-1))][j-1] );
    }
}

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

int main()
{
    while ( cin >> n )
    {
        if ( n == 0 )break;
        cin >> m;
        int start = 1;
        int tmp = 1;
        cnt.clear();
        for ( int i = 1; i <= n; i ++ )
        {
            cin >> a[i];
        }
        a[n+1] = 1<<28;
        for ( int i = 1; i <= n + 1; i ++ )
        {
            if ( a[i] > a[i-1] )
            {
                cnt.push_back( i - start );
                l[tmp] = start;
                r[tmp] = i - 1;
                for ( int j = start; j < i; j ++ )
                {
                    num[j] = tmp;
                }
                start = i;
                tmp ++;
            }
        }
        int a, b;
        rmq_init();
        for ( int i = 0; i < m; i ++ )
        {
            cin >> a >> b;
            if ( num[a] == num[b] )
            {
                cout << b - a + 1 << endl;
            }
            else
            {
                int ans = 0;
                ans = max ( r[num[a]] - a + 1, b - l[num[b]] + 1 );
                ans = max ( ans, rmq( num[a]+1, num[b]-1 ) );
                cout << ans << endl;
            }
        }
    }
    return 0;
}