poj 3368 Frequent values--RMQ

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原题链接：http://poj.org/problem?id=3368

题意：n个数，q个查询，每次查询，两个下标low，high，求在下标之间的数的频率的最大值。

分析：因为是求出频率的最大值，不放设f数组是频率，预先求出即可。对于任意的low，high，我们分为两部分，左部分连续相同的值，和剩下的右部分。左部分连续相同的个数很容易求出，右部分只需求出对应的f数组的最大值即可，右部分用RMQ来做。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#include<string>#include<stdio.h>#define INF 1000000000#define EPS 1e-6using namespace std;int n, q;int a[100005];int f[100005];int maxx[100005][20];void RMQ(){    for (int i = 1; i <= n; i++)        maxx[i][0] = f[i];    int k = log(double(n)) / log(2.0);    for (int j = 1; j <= k; j++)        for (int i = 1; i + (1 << j) - 1 <= n; i++)            maxx[i][j] = max(maxx[i][j - 1], maxx[i + (1 << (j - 1))][j - 1]);}int query(int low, int high){    if (low > high)        return 0;    int k = log(double(high - low + 1)) / log(2.0);    return max(maxx[low][k], maxx[high - (1 << k) + 1][k]);}int main(){    while (~scanf("%d", &n) && n)    {        a[0] = INF;        scanf("%d", &q);        for (int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            if (a[i] == a[i - 1])                f[i] = f[i - 1] + 1;            else                f[i] = 1;        }        RMQ();        int low, high;        for (int i = 0; i < q; i++)        {            scanf("%d%d", &low, &high);            int num = 0;            int t = low;            while (t <= high&&a[t] == a[t - 1])            {                t++;                num++;            }            printf("%d\n", max(num, query(t, high)));        }    }    return 0;}

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