POJ 3368 Frequent values (ST表)

Posted 2020-12-21 hhlya

题面

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.

题解

给你一个序列，然后有m次的查询，每次查询给你左右端点，问你区间内的最长连续序列是多长（元素全部相同）。区间询问，考虑st表，那么我们统计每个序列是以它为值的答案序列的第几项。然后因为是有序序列，所以查询的时候先二分出大于他的哪个元素下标，然后对前一个与右边界比较，如果大于那么就询问的区间长度就是答案吗，如果不是的话，那么我们就比较二分到的长度，和后半部分的答案，去取max，后半部分的答案可以用st表来维护（维护num值）。

代码实现

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const int inf=0x3f3f3f3f;
typedef pair<int,int> PII;
void check_min (int &a,int b) {a=min (a,b);}
void check_max (int &a,int b) {a=max (a,b);}
int n,m;
int data[maxn],num[maxn],dp[20][maxn];

inline void RMQ () {
  for (int i=1;i<=n;i++) dp[0][i]=num[i];
  int k=__lg (n+1);
  for (int j=1;j<=k;j++) 
   for (int i=1;i+(1<<j-1)<=n;i++) {
    dp[j][i]=max (dp[j-1][i],dp[j-1][i+(1<<(j-1))]);
   }
  return ;
}

int ask (int x,int y) {
  int k=__lg (y-x+1);
  return max (dp[k][x],dp[k][y-(1<<k)+1]);
}

int main () {
  while (scanf ("%d",&n)) {
    memset (num,0,sizeof (num));
    // for (int i=1;i<=n;i++) printf ("%d ",num[i]);
    // printf ("
");
    if (n==0) break;
    scanf ("%d",&m);
    for (int i=1;i<=n;i++) scanf ("%d",&data[i]);
    for (int i=1;i<=n;i++) {
      if (i==1||data[i]!=data[i-1]) num[i]=1;
      else num[i]=num[i-1]+1;
    }
    RMQ ();
    // for (int i=1;i<=n;i++) printf ("%d ",&num[i]);
    // printf ("
");
    while (m--) {
      int a,b,ans=1;
      scanf ("%d%d",&a,&b);
      int pos=upper_bound (data+1,data+1+n,data[a])-data;
      if (pos>b) ans=b-a+1;
      else {
        ans=pos-a;
        check_max (ans,ask (pos,b));
      }
      printf ("%d
",ans);
    }
  }
  return 0;
}