UVA 11235 Frequent values 非递减序列 l r范围内 出现最多的数字次数 RMQ

Frequent values

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

2007/2008 ACM International Collegiate Programming Contest 
University of Ulm Local Contest

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 10 102 31 105 100

Sample Output

143

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A naive algorithm may not run in time!

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=23846

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2176

题意：

输入一个n个元素的非减序列a[]，接着进行q次询问，每次询问输入两个数L, R， 问a[L]与a[R]之间相同元素的个数最多有多少个。

 

思路：  把每个相同的连续串合并到一起 分成多个段 段的数值一样   ： 用（a，b）表示有b个连续的a     用num[i]  left[pos]     right[pos]   分别表示位置pos所在的段的编号 和左右端点位置  用count[i]表示第i段的出现次数

对于查询 l r    如果 lr再同一段 则结果为r-l+1  

如果不在同一段 则为求这中间多个段内的最大的count[i]值（RMQ）   以及l r所在的段的连续个数的最大值  

具体看代码

 

#include<stdio.h>#include<string.h>const int size=100111;int value[size],count[size],num[size],left[size],right[size];int d[size][50];int mmax(int a,int b){if(a>b) return a;return b;}void  RMQ_init(int n){int i,j;       for(i=1;i<=n;i++) d[i][0]=count[i];   for(j=1;(1<<j)<=n;j++)   for(i=1;i+j-1<=n;i++)   d[i][j]=mmax(d[i][j-1],d[i+(1<<(j-1))][j-1]);}int RMQ(int l,int r){int k=0;while(1<<(k+1)<=r-l+1) k++;return mmax(d[l][k],d[r-(1<<k)+1][k]);}int main(){int n,m,i,j,k,last;while(scanf("%d",&n)!=EOF){if(!n) break;scanf("%d",&m);last=-99999999;int id=0;//id表示段 有效id从1开始int cnt=0;//cnt表示出现次数int l=0,r=0;        for(i=1;i<=n;i++){scanf("%d",&k);if(k!=last){r=i-1;for(j=l;j<=r;j++){                       num[j]=id;   left[j]=l;   right[j]=r;}if(cnt==0) cnt=1;                count[id]=cnt;id++;cnt=1;l=i;last=k;}else{cnt++;last=k;}}for(i=l;i<=n;i++){                       num[i]=id;   left[i]=l;   right[i]=n;}count[id]=cnt;//for(i=1;i<=n;i++)//{//printf("所在段num[i]=%d left[i]=%d right[i]=%d count[id]=%d\n"//,num[i],left[i],right[i],count[num[i]]);//}RMQ_init(id);while(m--){int mid,l,r,id1,id2,l_max,r_max,ans;scanf("%d %d",&l,&r);id1=num[l];id2=num[r];if(id1==id2)  {printf("%d\n",r-l+1);continue;}            l_max=right[l]-l+1; r_max=r-left[r]+1;ans=l_max>r_max?l_max:r_max;if(id2-id1==1){                   printf("%d\n",ans);   continue;}mid=RMQ(id1+1,id2-1);ans=mmax(ans,mid);printf("%d\n",ans);}}return 0;}