POJ 3261 Milk Patterns

题目大意:

给N个数, 和一个正整数K, 求出在这串数中出现过至少K次的子串的最大长度, 这里对出现的字串计数时可以重叠.

 

简要分析:

二分答案M, 然后从左到右算一遍长为M的子串的哈希值, 丢到map里面进行计数, 时间复杂度O(NlogN).

 

代码实现:

View Code

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 typedef unsigned long long hash_t;
 8 const int MAX_N = 20000, SEED = 99991;
 9 int n, k, s[MAX_N];
10 hash_t mypow[MAX_N];
11 
12 struct hash_map_t {
13     static const int MOD = 196613;
14     int cnt, begin[MOD], val[MAX_N], next[MAX_N];
15     hash_t end[MAX_N];
16     void clear() { cnt = 0, memset(begin, -1, sizeof(begin)); }
17     hash_map_t() { cnt = 0, memset(begin, -1, sizeof(begin)); }
18     int push(hash_t x) {
19         int s = x % MOD;
20         for (int now = begin[s]; now != -1; now = next[now])
21             if (end[now] == x) {
22                 val[now] ++;
23                 return val[now];
24             }
25         next[cnt] = begin[s];
26         begin[s] = cnt;
27         end[cnt] = x;
28         val[cnt ++] = 1;
29         return 1;
30     }
31 } cnt;
32 
33 bool check(int len) {
34     cnt.clear();
35     hash_t t = 0ULL;
36     for (int i = 0; i < len; i ++) t = t * SEED + s[i];
37     cnt.push(t);
38     for (int i = 1; i < n - len + 1; i ++) {
39         t = (t - s[i - 1] * mypow[len - 1]) * SEED + s[i + len - 1];
40         if (cnt.push(t) >= k) return 1;
41     }
42     return 0;
43 }
44 
45 int main() {
46     mypow[0] = 1ULL;
47     for (int i = 1; i < MAX_N; i ++) mypow[i] = mypow[i - 1] * SEED;
48     while (scanf("%d%d", &n, &k) != EOF) {
49         for (int i = 0; i < n; i ++) scanf("%d", &s[i]);
50         int lb = 0, rb = n;
51         while (lb + 1 < rb) {
52             int mid = (lb + rb) >> 1;
53             if (check(mid)) lb = mid;
54             else rb = mid;
55         }
56         printf("%d\n", lb);
57     }
58     return 0;
59 }