poj Frequent values 3368

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 17832 Accepted: 6427

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 10 102 31 105 100

这个题我是被自己感动到了，没想到自己做的第二个RMQ的题可以一次就过，编译也是一次过，太感动了！

哈哈！

题意是：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

说一下如何做

正如《算法竞赛入门经典（训练指南）》198页讲的，

这里我们因为是按非降序排列的，那也就是说相同的数字一定是在一起的

这里我们就用相同的数字为一段来描述

那么说例题中我们就有2 4 1 3分别代表四段

我们用a[]数组来存给出的值，

下面说说我们用到的数组的意义

value[]: 代表第i段的具体数值

coun[]: 代表第i段出现的次数

lef[] : 代表第i段的左端点的位置

righ[]: : 代表第i段的右端点的位置

num[] 数组稍有不同 他指的是第p个位置上为第几段

这样我们只要找num[L]+1, num[R]-1段中的最大值与righ[L]-L+1, R-lef[R]+1,三个中的最大值就可以了，

当然特殊情况也要有考虑

大致可分为三种：

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int inf = INT_MAX;int value[300005], coun[300005];int num[300005], lef[300005], righ[300005];int a[300005], b[300005][20];int top;void inti(){    a[0] = inf;    top = 0;}void RMQ(){    int i, j;    for ( i = 1;i <= top; i++ )        b[i][0] = coun[i];    for ( j = 1;(1<<j) <= top; j++ )    {        for ( i = 1;i+(1<<(j-1)) <= top; i++ )            b[i][j] = max(b[i][j-1], b[i+(1<<(j-1))][j-1]);    }}int RMQ_check(int y, int x){    int nb = 0;    while ( (1<<nb) <= y-x+1 )        nb++;    return max(b[y-(1<<(nb-1))+1][nb-1], b[x][nb-1]);}int main(){    int i, n, m;    int L, R;    while ( ~scanf ( "%d", &n )&&n != 0 )    {        scanf ( "%d", &m );        inti();        for ( i = 1;i <= n; i++ )        {            scanf ( "%d", &a[i] );            if ( a[i] != a[i-1] )            {                righ[top] = i-1;                top++;                lef[top] = i;                value[top] = a[i];                coun[top] = 0;            }            num[i] = top;            coun[top]++;        }        righ[top] = n;        RMQ();        for ( i = 0;i < m; i++ )        {            scanf ( "%d %d", &L, &R );            if ( num[L] == num[R] )                printf ( "%d\n", R-L+1 );            else if ( num[R]-num[L] == 1 )                printf ( "%d\n", max(righ[num[L]]-L+1, R-lef[num[R]]+1) );            else            {                int ma1 = righ[num[L]]-L+1;                int ma2 =  R-lef[num[R]]+1;                int ma3 = RMQ_check(num[R]-1, num[L]+1);                printf ( "%d\n", max(ma1, max(ma2, ma3)));            }        }    }    return 0;}

代码菜鸟，如有错误，请多包涵！！！

如有帮助记得支持我一下，谢谢！！！