poj 3368（RMQ简单应用）

Frequent values

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13317 Accepted: 4894

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

Source

Ulm Local 2007

详细参考刘汝佳的算法训练指南第198页，看了就懂了。

AC代码：

#include<stdio.h>#include<iostream>#include<stdio.h>#include<algorithm>using namespace std;int a[100010];int val[100010],counter[100010];int num[100010],Left[100010],Right[100010];int d[100100][110];int n,q;void RMQ_init(){    int i,j,k;    int tmp;    val[1]=a[1];    k=num[1]=Left[1]=1;    for(i=2;i<=n;i++){        if(a[i]!=val[k]){            Right[k]=i-1;            counter[k]=Right[k]-Left[k]+1;            k++;            Left[k]=i;            val[k]=a[i];        }        num[i]=k;    }    Right[k]=n;    counter[k]=Right[k]-Left[k]+1;    for(i=1;i<=k;i++)        d[i][0]=counter[i];    for(j=1;(1<<j)<=k;j++)    for(i=1;i+(1<<j)-1<=k;i++){        d[i][j]=max(d[i][j-1],d[i+(1<<(j-1))][j-1]);    }}int RMQ(int x,int y){    int k=0;    while((1<<(k+1))<=y-x+1) k++;    return max(d[x][k],d[y-(1<<k)+1][k]);}int main(){    while(scanf("%d",&n)!=EOF){        if(n==0)            break;        scanf("%d",&q);        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        RMQ_init();        while(q--){            int x,y;            scanf("%d%d",&x,&y);            if(num[x]==num[y]){                printf("%d\n",y-x+1);            }            else{                int ans;                ans=max(Right[num[x]]-x+1,y-Left[num[y]]+1);                if(num[x]+1<=num[y]-1){                    ans=max(ans,RMQ(num[x]+1,num[y]-1));                }                printf("%d\n",ans);            }        }    }    return 0;}