POJ 3368 Frequent values

　　简单的RMQ  约等于模板题.....第一次敲 给自己留个参考

　　

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

#define Max(a,b) a > b ? a : b

using namespace std;

int Num[100010];

int MaxNum[100010][15];

int value[100010],Count[100010],Left[100010],Right[100010];

int divide(int Num,int s,int e)
{
    int mid = (s+e)/2;
    if(Num == value[mid])
        return mid;
    else if(Num < value[mid])
        return divide(Num,s,mid-1);
    return divide(Num,mid+1,e);
}

int RMQ(int L,int R)
{
    if(L > R)
        return -1;
    else if(L == R)
        return Count[L];
    int k = 0;
    while( (1<<(k+1)) <= R-L+1 )
        ++k;
    return Max(MaxNum[L][k],MaxNum[R-(1<<k)+1][k]);
}

void RMQ_init(int n)
{
    int i,j;
    for(i = 1; i <= n; ++i)
    {
        MaxNum[i][0] = Count[i];
    }

    for(j = 1; (1<<j) <= n; ++j)
    {
        for(i = 1; i + j - 1 <= n; ++i)
        {
            MaxNum[i][j] = Max(MaxNum[i][j-1],MaxNum[i+(1<<(j-1))][j-1]);
        }
    }
}

int main()
{
    int n,m,i,top,last,L,R;
    while(scanf("%d",&n) && n)
    {
        scanf("%d",&m);

        top = 1;
        last = -200010;
        Left[0] = 1;

        for(i = 1; i <= n; ++i)
        {
            scanf("%d",&Num[i]);
            if(Num[i] != last)
            {
                last = Num[i];
                value[top] = Num[i];
                Left[top] = i;
                Right[top-1] = i-1;
                Count[top-1] = Right[top-1] - Left[top-1] + 1;
                ++top;
            }
        }

        RMQ_init(top-1);
        while(m--)
        {
            scanf("%d %d",&L,&R);
            int LI = divide(Num[L],1,top);
            int RI = divide(Num[R],1,top);

            if(LI == RI)
                printf("%d
",R-L+1);
            else
            {
                int MaxNum = Max(Right[LI] - L + 1,R - Left[RI] + 1);
                int temp = RMQ(LI+1,RI-1);
                printf("%d
", (Max(temp,MaxNum)) );
            }
        }
    }
    return 0;
}