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 = ;
 const int MAXM = ;
 const double eps = 1e-;

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

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

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

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