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RMQ [HDU 1806] Frequent values
Frequent values
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1146 Accepted Submission(s): 415
Problem Description
You are given a sequence of n integers a1 , a2 , ... , an 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 ai , ... , aj .
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 , ... , an(-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+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.
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
Hint
A naive algorithm may not run in time!
Source
HDOJ 2007 Summer Exercise(1)
分成三段、中间RMQ、然后求最大值即可
#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define N 100010int n,m;int a[N];int dp[N][20];int id[N];int len[N],l[N],r[N];void init(){ int i,j; for(i=1;i<=n;i++) { dp[i][0]=len[i]; } int k=(int)(log((double)n)/log(2.0)); for(j=1;j<=k;j++) { for(i=1;i+(1<<j)-1<=n;i++) { dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } }}int query(int i,int j){ int k=(int)(log((double)(j-i+1))/log(2.0)); int res=max(dp[i][k],dp[j-(1<<k)+1][k]); return res;}int main(){ int i,pos; while(scanf("%d",&n),n) { pos=0; scanf("%d",&m); for(i=1;i<=n;i++) { scanf("%d",&a[i]); if(a[i]!=a[i-1]) { pos++; l[pos]=i; } id[i]=pos; r[pos]=i; len[pos]=r[pos]-l[pos]+1; } n=pos; init(); int x,y,xx,yy,ans1,ans2,ans3; while(m--) { int x,y; scanf("%d%d",&x,&y); xx=id[x]; yy=id[y]; if(xx==yy) ans1=ans2=y-x+1; //特殊情况、当x和y在同一个区间、答案是y-x+1 else { ans1=r[xx]-x+1; ans2=y-l[yy]+1; } ans3=0; if(xx+1<=yy-1)ans3=query(xx+1,yy-1); printf("%d\n",max(max(ans1,ans2),ans3)); } } return 0;}
RMQ [HDU 1806] Frequent values
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