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 #include2 #include 3 #include 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 }