SPOJ 694&&SPOJ705: Distinct Substrings
DISUBSTR - Distinct Substrings
题意：
询问有多少不同的子串。
思路：
后缀数组或者SAM。
首先求出后缀数组，然后从对于一个后缀，它有n-sa[i]-1个前缀，其中有height[rnk[i]]个被rnk[i]-1的后缀算了。所以再减去height[rnk[i]]即可。
代码：
换了板子。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 7 using namespace std; 8 9 const int N = 1010; 10 char s[N]; 11 int t1[N],t2[N],sa[N],height[N],rnk[N],c[N]; 12 int n; 13 14 void get_sa(int m) { 15 int *x = t1,*y = t2,i,p; 16 for (i=1; i<=m; ++i) c[i] = 0; 17 for (i=1; i<=n; ++i) x[i]=s[i],c[x[i]]++; 18 for (i=1; i<=m; ++i) c[i] += c[i-1]; 19 for (i=n; i>=1; --i) sa[c[ x[i] ]--] = i; 20 for (int k=1; k<=n; k<<=1) { 21 p = 0; 22 for (i=n-k+1; i<=n; ++i) y[++p] = i; 23 for (i=1; i<=n; ++i) if (sa[i]>k) y[++p] = sa[i]-k; 24 for (i=1; i<=m; ++i) c[i] = 0; 25 for (i=1; i<=n; ++i) c[ x[y[i]] ]++; 26 for (i=1; i<=m; ++i) c[i] += c[i-1]; 27 for (i=n; i>=1; --i) sa[c[ x[y[i]] ]--] = y[i]; 28 swap(x,y); 29 p = 2; 30 x[sa[1]] = 1; 31 for (i=2; i<=n; ++i) 32 x[sa[i]] = y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k] ? p-1 : p++; 33 if (p > n) break; 34 m = p; 35 } 36 } 37 void get_height() { 38 for (int i=1; i<=n; ++i) rnk[sa[i]] = i; 39 int k = 0; 40 height[1] = 0; 41 for (int i=1; i<=n; ++i) { 42 if (rnk[i]==1) continue; 43 if (k) k--; 44 int j = sa[rnk[i]-1]; 45 while (i+k<=n && j+k<=n && s[i+k]==s[j+k]) 46 k++; 47 height[rnk[i]] = k; 48 } 49 } 50 void get_ans() { 51 long long ans = 0; 52 for (int i=1; i<=n; ++i) { 53 ans += max(n-i+1-height[rnk[i]],0); 54 // ans += max(n-sa[i]-height[i],0); 55 } 56 cout << ans << '\n'; 57 } 58 int main() { 59 int T; 60 scanf("%d",&T); 61 while (T--) { 62 scanf("%s",s+1); 63 n = strlen(s+1); 64 get_sa(130); 65 get_height(); 66 get_ans(); 67 } 68 return 0; 69 }
原来的
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 7 using namespace std; 8 9 const int N = 1010; 10 char s[N]; 11 int t1[N],t2[N],sa[N],height[N],rnk[N],c[N]; 12 int n; 13 14 void get_sa(int m) { 15 int *x = t1,*y = t2; 16 for (int i=0; i<m; ++i) c[i] = 0; 17 for (int i=0; i<n; ++i) c[x[i] = s[i]]++; 18 for (int i=1; i<m; ++i) c[i] += c[i-1]; 19 for (int i=n-1; i>=0; --i) sa[--c[x[i]]] = i; 20 for (int k=1; k<=n; k<<=1) { 21 int p = 0; 22 for (int i=n-k; i<n; ++i) y[p++] = i; 23 for (int i=0; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k; 24 for (int i=0; i<m; ++i) c[i] = 0; 25 for (int i=0; i<n; ++i) c[x[y[i]]]++; 26 for (int i=1; i<m; ++i) c[i] += c[i-1]; 27 for (int i=n-1; i>=0; --i) sa[--c[x[y[i]]]] = y[i]; 28 swap(x,y); 29 p = 1; 30 x[sa[0]] = 0; 31 for (int i=1; i<n; ++i) 32 x[sa[i]] = (y[sa[i-1]]==y[sa[i]] && sa[i-1]+k<n && sa[i]+k<n && 33 y[sa[i-1]+k]==y[sa[i]+k]) ? p-1 : p++; 34 35 if (p >= n) break; 36 m = p; 37 } 38 } 39 void get_height() { 40 for (int i=0; i<n; ++i) rnk[sa[i]] = i; 41 int k = 0; 42 height[0] = 0; 43 for (int i=0; i<n; ++i) { 44 if (!rnk[i]) continue; 45 if (k) k--; 46 int j = sa[rnk[i]-1]; 47 while (i+k < n && j+k < n && s[i+k]==s[j+k]) k++; 48 height[rnk[i]] = k; 49 } 50 } 51 void get_ans() { 52 long long ans = 0; 53 for (int i=0; i<n; ++i) { 54 ans += max(n-i-height[rnk[i]],0); 55 // ans += max(n-sa[i]-height[i],0); 56 } 57 cout << ans << '\n'; 58 } 59 int main() { 60 int T; 61 scanf("%d",&T); 62 while (T--) { 63 scanf("%s",s); 64 n = strlen(s); 65 get_sa(130); 66 get_height(); 67 get_ans(); 68 } 69 return 0; 70 }
