SPOJ 694 (后缀数组) Distinct Substrings

将所有后缀按照字典序排序后,每新加进来一个后缀,它将产生n - sa[i]个前缀。这里和小罗论文里边有点不太一样。

height[i]为和字典序前一个的LCP,所以还要减去,最终累计n - sa[i] - height[i]即可。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const int maxn = 100000 + 10;
 7 char s[maxn];
 8 int sa[maxn], rank[maxn], height[maxn];
 9 int t[maxn], t2[maxn], c[maxn];
10 int n;
11 
12 void build_sa(int n, int m)
13 {
14     int i, *x = t, *y = t2;
15     for(i = 0; i < m; i++) c[i] = 0;
16     for(i = 0; i < n; i++) c[x[i] = s[i]]++;
17     for(i = 1; i < m; i++) c[i] += c[i - 1];
18     for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
19     for(int k = 1; k <= n; k <<= 1)
20     {
21         int p = 0;
22         for(i = n - k; i < n; i++) y[p++] = i;
23         for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
24         for(i = 0; i < m; i++) c[i] = 0;
25         for(i = 0; i < n; i++) c[x[y[i]]]++;
26         for(i = 1; i < m; i++) c[i] += c[i - 1];
27         for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
28         swap(x, y);
29         p = 1; x[sa[0]] = 0;
30         for(i = 1; i < n; i++)
31             x[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k] ? p - 1 : p++;
32         if(p >= n) break;
33         m = p;
34     }
35 }
36 
37 void build_height()
38 {
39     int k = 0;
40     for(int i = 1; i <= n; i++) rank[sa[i]] = i;
41     for(int i = 0; i < n; i++)
42     {
43         if(k) k--;
44         int j = sa[rank[i] - 1];
45         while(s[i + k] == s[j + k]) k++;
46         height[rank[i]] = k;
47     }
48 }
49 
50 int main()
51 {
52     //freopen("in.txt", "r", stdin);
53 
54     int T; scanf("%d", &T);
55     while(T--)
56     {
57         scanf("%s", s);
58         n = strlen(s);
59         memset(sa, 0, sizeof(sa));
60         build_sa(n + 1, 256);
61         build_height();
62 
63         int ans = 0;
64         for(int i = 1; i <= n; i++)
65             ans += n - sa[i] - height[i];
66         printf("%d\n", ans);
67     }
68 
69     return 0;
70 }
代码君

 

posted @ 2015-04-24 09:23  AOQNRMGYXLMV  阅读(198)  评论(0编辑  收藏  举报