POJ 2774 Long Long Message

最新推荐文章于 2019-03-31 14:23:32 发布

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最新推荐文章于 2019-03-31 14:23:32 发布

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原文链接：http://www.cnblogs.com/zcwwzdjn/archive/2012/03/09/2388373.html

本文介绍了一种算法，用于求解两个长度不超过100000的字符串之间的最长公共子串问题，通过二分搜索和哈希映射实现O(NlogN)的时间复杂度。

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题目大意:

求两个长度不超过100000的字符串的最长公共子串, 注意是子串而非子序列, 所以是连续的.

简要分析:

二分答案M, 设两个串分别为A和B, 从左到右枚举一遍A的长为M的子串, 算出哈希值丢到map里面, 再从左到右枚举一遍B的长为M的子串, 算出哈希再到map里面找. 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 = 100000, SEED = 99991;
 9 char a[MAX_N + 1], b[MAX_N + 1];
10 int lena, lenb;
11 hash_t mypow[MAX_N];
12 
13 struct hash_map_t {
14     static const int MOD = 196613;
15     int cnt, begin[MOD], next[MAX_N];
16     hash_t end[MAX_N];
17     void clear() { cnt = 0, memset(begin, -1, sizeof(begin)); }
18     hash_map_t() { cnt = 0, memset(begin, -1, sizeof(begin)); }
19     void push(hash_t x) {
20         int s = x % MOD;
21         for (int now = begin[s]; now != -1; now = next[now])
22             if (end[now] == x) return;
23         next[cnt] = begin[s];
24         begin[s] = cnt;
25         end[cnt ++] = x;
26     }
27     bool ask(hash_t x) {
28         int s = x % MOD;
29         for (int now = begin[s]; now != -1; now = next[now])
30             if (end[now] == x) return 1;
31         return 0;
32     }
33 } hash;
34 
35 bool check(int len) {
36     hash.clear();
37     hash_t t = 0ULL;
38     for (int i = 0; i < len; i ++) t = t * SEED + a[i];
39     hash.push(t);
40     for (int i = 1; i < lena - len + 1; i ++) {
41         t = (t - mypow[len - 1] * a[i - 1]) * SEED + a[i + len - 1];
42         hash.push(t);
43     }
44     t = 0ULL;
45     for (int i = 0; i < len; i ++) t = t * SEED + b[i];
46     if (hash.ask(t)) return 1;
47     for (int i = 1; i < lenb - len + 1; i ++) {
48         t = (t - mypow[len - 1] * b[i - 1]) * SEED + b[i + len - 1];
49         if (hash.ask(t)) return 1;
50     }
51     return 0;
52 }
53 
54 int main() {
55     mypow[0] = 1ULL;
56     for (int i = 1; i < MAX_N; i ++) mypow[i] = mypow[i - 1] * SEED;
57     while (scanf("%s%s", a, b) != EOF) {
58         lena = strlen(a), lenb = strlen(b);
59         int lb = 0, rb = min(lena, lenb) + 1;
60         while (lb + 1 < rb) {
61             int mid = (lb + rb) >> 1;
62             if (check(mid)) lb = mid;
63             else rb = mid;
64         }
65         printf("%d\n", lb);
66     }
67     return 0;
68 }