PKU 3368（线段树 + 离散化）

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5869 Accepted: 2062

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

Sample Output

143

//ai ≤ ai+1   题目已经限制了这个，不然难度就上升了#include <iostream>using namespace std;#define N 100010#define max(a,b) (a>b?a:b)struct LineTree {int l,r,len,mid;int a,b,MAX;   //a记录左节点的值   b记录右节点的值  MAX记录该区间最长的长度}tree[N*3];int res[N];int len;int val[N];void Build(int root,int a,int b){tree[root].l = a;tree[root].r = b;tree[root].mid = (a+b)>>1;tree[root].len = b-a;tree[root].a = res[a];tree[root].b = res[b];if(tree[root].len == 1){tree[root].MAX = res[b] - res[a];return;}Build(root<<1,a,tree[root].mid);Build((root<<1)+1,tree[root].mid,b);tree[root].MAX = max(tree[root<<1].MAX , tree[(root<<1)+1].MAX);}int Search(int root,int a,int b){if(tree[root].a == a && tree[root].b == b){return tree[root].MAX;}if(tree[root].len == 1) return b-a;if(b <= tree[root<<1].b) return Search(root<<1,a,b);else if(tree[(root<<1)+1].a <= a) return Search((root<<1)+1,a,b);else{return max(Search(root<<1,a,tree[root<<1].b),Search((root<<1)+1,tree[(root<<1)+1].a,b));}}int main(){int n,m,i,a,b;while (scanf("%d",&n),n){scanf("%d",&m);for(i=1;i<=n;i++)scanf("%d",&val[i]);len=2;res[1] = 1;for(i=2;i<=n;i++){if(val[i] != val[i-1])res[len++] = i;}res[len] = n+1;Build(1,1,len);while (m--){scanf("%d%d",&a,&b);printf("%d/n",Search(1,a,b+1));}}return 0;}