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 ", ans); 67 } 68 69 return 0; 70 }