UVa11235 FrequentValues（RMQ）

Problem F: Frequent values

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 Specification

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 Specification

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

题目大意：
给一个非降序排列的整数数组a，你的任务是对于一系列询问(i, j)，回答ai,ai+1...aj中次数出现最多的值所出现的次数。

分析：
由于数列是非降序的，所以所有相等的数都会聚集在一起。这样我们就可以把整个数组进行编码。如-1,1,1,2,2,2,4就可以编码成(-1,1),(1,2),(2,3),(4,1)表示（a，b）数组中的a连续出现了b次。用num[i]表示原数组下表是i的数在编码后的第num[i]段。left[i],right[i]表示第i段的左边界和右边界，用coun[i]表示第i段有conu[i]个相同的数。这样的话每次查询(L, R)就只要计算(right[L]-L+1),(R-left[R]+1)和RMQ(num[L]+1, num[R]-1)这三个值的最大值就可以了。
其中，RMQ是对coun数组进行取件查询的结果。
特殊的，如果L和R在同一个区间内的话，那么结果就是（R-L+1）

详见代码：

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson k<<1, L, mid
#define rson k<<1|1, mid+1, R
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define FOPENIN(IN) freopen(IN, "r", stdin)
#define FOPENOUT(OUT) freopen(OUT, "w", stdout)

template<class T> T CMP_MIN(T a, T b) { return a < b; }
template<class T> T CMP_MAX(T a, T b) { return a > b; }
template<class T> T MAX(T a, T b) { return a > b ? a : b; }
template<class T> T MIN(T a, T b) { return a < b ? a : b; }
template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }

//typedef __int64 LL;
typedef long long LL;
const int MAXN = 100005;
const int MAXM = 100005;
const double eps = 1e-12;

int num[MAXN], coun[MAXN], Left[MAXN], Right[MAXN];
int n, q, a, last, tot;
int DP[MAXN][20];

void init_RMQ()
{
    mem0(DP);
    for(int i=1;i<=tot;i++) DP[i][0] = coun[i];
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)<=tot;i++)
        {
            DP[i][j] = max(DP[i][j-1], DP[i+(1<<(j-1))][j-1]);
        }
    }
}

int RMQ(int L, int R)
{
    if(L > R) return 0;
    int k = 0;
    while((1<<(1+k)) <= R-L+1) k++;
    return max(DP[L][k], DP[R-(1<<k)+1][k]);
}

int main()
{
//    FOPENIN("in.txt");
//    FOPENOUT("out.txt");
    while(~scanf("%d", &n) && n)
    {
        scanf("%d", &q);
        tot = 0; mem0(Left); mem0(Right); mem0(coun);
        for(int i=1;i<=n;i++)
        {
            scanf("%d", &a);
            if(i==1) { ++tot;   last=a;  Left[tot] = 1; }
            if(last == a) { num[i]=tot; coun[tot]++; Right[tot]++; }
            else { num[i]=++tot; coun[tot]++; Left[tot]=Right[tot]=i; last=a; }
        }
        init_RMQ();
        int l, r;
        for(int i=0;i<q;i++)
        {
            scanf("%d%d", &l, &r);
            if(num[l] == num[r]) { printf("%d
", r-l+1);  continue; }
            printf("%d
", max( RMQ(num[l]+1, num[r]-1), max( Right[num[l]]-l+1, r-Left[num[r]]+1 ) ) );
        }
    }
    return 0;
}