hdu1403 后缀数组入门题

//1403

思路:字符串的任何一个子串都是这个字符串的某个后缀的前缀,则求A和B的最长公共子串等价于求A的后缀和B的后缀的最长公共前缀的最大值。

做法:将第二个字符串写在第一个字符串后面,中间用一个没有出现过的字符隔开,再求这个新的字符串的后缀数组。

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 char str[maxn];
 5 int int_str[maxn];
 6 int len;
 7 int str_rank[maxn], s_r[maxn], rank_str[maxn], a[maxn], b[maxn], h_rank[maxn];
 8 
 9 int Count[maxn];
10 void radix(int *str, int *s_r, int *str_rank, int len, int m)
11 {
12     int i;
13     memset(Count, 0, sizeof(Count));
14     for(i = 0; i < len; ++i)
15         ++Count[str[s_r[i]]];
16     for(i = 1; i <= m; ++i)
17         Count[i] += Count[i - 1];
18     for(i = len - 1; i >= 0; --i) {
19         str_rank[--Count[str[s_r[i]]]] = s_r[i];
20     }
21 }
22 
23 void suffix_array(int *str, int *str_rank, int len, int m)
24 {
25     int i;
26     for(i = 0; i < len; ++i)
27         s_r[i] = i;
28     radix(str, s_r, str_rank, len, m);
29     rank_str[str_rank[0]] = 0;
30     for(i = 1; i < len; ++i) {
31         rank_str[str_rank[i]] = rank_str[str_rank[i - 1]] + (str[str_rank[i - 1]] != str[str_rank[i]]);
32     }
33     int j;
34     for(i = 0; (1 << i) < len; ++i) {
35         for(j = 0; j < len; ++j) {
36             a[j] = rank_str[j] + 1;
37             b[j] = ((1 << i) + j < len)? rank_str[(1 << i) + j] + 1: 0;
38             str_rank[j] = j;
39         }
40         radix(b, str_rank, s_r, len, len);
41         radix(a, s_r, str_rank, len, len);
42         rank_str[str_rank[0]] = 0;
43         for(j = 1; j < len; ++j) {
44             rank_str[str_rank[j]] = rank_str[str_rank[j - 1]] + (a[str_rank[j - 1]] != a[str_rank[j]] || b[str_rank[j - 1]] != b[str_rank[j]]);
45         }
46     }
47 }
48 
49 void cal_h(int *str, int *str_rank, int *h_rank, int len)
50 {
51     int i, k = 0;
52     for(i = 0; i < len; ++i) {
53         rank_str[str_rank[i]] = i;
54     }
55     for(i = 0; i < len; ++i) {
56         k = (k == 0? 0: k - 1);
57         if(rank_str[i] != 0) {
58             while(str[i + k] == str[str_rank[rank_str[i] - 1] + k])
59                 ++k;
60         }
61         h_rank[rank_str[i]] = k;
62     }
63 }
64 
65 
66 int main()
67 {
68     while(scanf("%s", str) != EOF) {
69         len = strlen(str);
70         int len1 = len;
71         str[len] = 'a' - 1;
72         scanf("%s", str + len + 1);
73         len = strlen(str);
74         int i;
75         for(i = 0; i < len; ++i)
76             int_str[i] = str[i];
77         suffix_array(int_str, str_rank, len, 200);
78         cal_h(int_str, str_rank, h_rank, len);
79 
80 //        for(i = 0; i < len; ++i) {
81 //            printf("%d: %s %d
", i + 1, str + str_rank[i], h_rank[i]);
82 //        }
83 
84         int res = 0;
85         for(i = 1; i < len; ++i) {
86             if(res < h_rank[i] && ((str_rank[i] < len1 && str_rank[i - 1] > len1) || (str_rank[i] > len1 && str_rank[i - 1] < len1)))
87                 res = h_rank[i];
88         }
89         printf("%d
", res);
90     }
91 
92 }
原文地址:https://www.cnblogs.com/AC-Phoenix/p/4455043.html