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  • POJ3261 Milk Patterns —— 后缀数组 出现k次且可重叠的最长子串

    题目链接:https://vjudge.net/problem/POJ-3261

    Milk Patterns
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 17157   Accepted: 7592
    Case Time Limit: 2000MS

    Description

    Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

    To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

    Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.

    Input

    Line 1: Two space-separated integers: N and K 
    Lines 2..N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.

    Output

    Line 1: One integer, the length of the longest pattern which occurs at least K times

    Sample Input

    8 2
    1
    2
    3
    2
    3
    2
    3
    1

    Sample Output

    4

    Source

    题意:

    给出一个字符串(数字串),问至少出现了k次并且可以重叠的最长子串的长度。

    题解:

    1.POJ1743 Musical Theme 此题的加强版。

    2.此题要求至少出现了k次,那么相应地在每一组内统计:是否存在一个长度不小于mid(二分时的mid),并且出现了至少k次的公共前缀。

    代码如下:

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <cstring>
      4 #include <algorithm>
      5 #include <vector>
      6 #include <cmath>
      7 #include <queue>
      8 #include <stack>
      9 #include <map>
     10 #include <string>
     11 #include <set>
     12 using namespace std;
     13 typedef long long LL;
     14 const double EPS = 1e-6;
     15 const int INF = 2e9;
     16 const LL LNF = 9e18;
     17 const int MOD = 1e5;
     18 const int MAXN = 1e6+10;
     19 
     20 bool cmp(int *r, int a, int b, int l)
     21 {
     22     return r[a]==r[b] && r[a+l]==r[b+l];
     23 }
     24 
     25 int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];
     26 int t1[MAXN], t2[MAXN], c[MAXN];
     27 void DA(int str[], int sa[], int Rank[], int height[], int n, int m)
     28 {
     29     n++;
     30     int i, j, p, *x = t1, *y = t2;
     31     for(i = 0; i<m; i++) c[i] = 0;
     32     for(i = 0; i<n; i++) c[x[i] = str[i]]++;
     33     for(i = 1; i<m; i++) c[i] += c[i-1];
     34     for(i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
     35     for(j = 1; j<=n; j <<= 1)
     36     {
     37         p = 0;
     38         for(i = n-j; i<n; i++) y[p++] = i;
     39         for(i = 0; i<n; i++) if(sa[i]>=j) y[p++] = sa[i]-j;
     40 
     41         for(i = 0; i<m; i++) c[i] = 0;
     42         for(i = 0; i<n; i++) c[x[y[i]]]++;
     43         for(i = 1; i<m; i++) c[i] += c[i-1];
     44         for(i = n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];
     45 
     46         swap(x, y);
     47         p = 1; x[sa[0]] = 0;
     48         for(i = 1; i<n; i++)
     49             x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
     50         if(p>=n) break;
     51         m = p;
     52     }
     53 
     54     int k = 0;
     55     n--;
     56     for(i = 0; i<=n; i++) Rank[sa[i]] = i;
     57     for(i = 0; i<n; i++)
     58     {
     59         if(k) k--;
     60         j = sa[Rank[i]-1];
     61         while(str[i+k]==str[j+k]) k++;
     62         height[Rank[i]] = k;
     63     }
     64 }
     65 
     66 bool test(int mid, int n, int k)
     67 {
     68     int cnt = 1;
     69     for(int i = 2; i<=n; i++)
     70     {
     71         if(height[i]<mid)
     72             cnt = 1;
     73         else
     74         {
     75             cnt++;
     76             if(cnt==k) return true;
     77         }
     78     }
     79     return false;
     80 }
     81 
     82 int main()
     83 {
     84     int n, k;
     85     while(scanf("%d%d",&n,&k)!=EOF)
     86     {
     87         for(int i = 0; i<n; i++) scanf("%d", &r[i]);
     88         r[n] = 0;
     89         DA(r, sa, Rank, height, n, 1e6+1);
     90         int l = 0, r = n/2;
     91         while(l<=r)
     92         {
     93             int mid = (l+r)>>1;
     94             if(test(mid, n, k))
     95                 l = mid + 1;
     96             else
     97                 r = mid - 1;
     98         }
     99         printf("%d
    ", r);
    100     }
    101 }
    View Code
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  • 原文地址:https://www.cnblogs.com/DOLFAMINGO/p/8466900.html
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