SPOJ 694&&SPOJ705: Distinct Substrings

DISUBSTR - Distinct Substrings

链接

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。

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<cmath>

 using namespace std;

 const int N = ;
 char s[N];
 int t1[N],t2[N],sa[N],height[N],rnk[N],c[N];
 int n;

 void get_sa(int m) {
     int *x = t1,*y = t2,i,p;
     for (i=; i<=m; ++i) c[i] = ;
     for (i=; i<=n; ++i) x[i]=s[i],c[x[i]]++;
     for (i=; i<=m; ++i) c[i] += c[i-];
     for (i=n; i>=; --i) sa[c[ x[i] ]--] = i;
     for (int k=; k<=n; k<<=) {
         p = ;
         for (i=n-k+; i<=n; ++i) y[++p] = i;
         for (i=; i<=n; ++i) if (sa[i]>k) y[++p] = sa[i]-k;
         for (i=; i<=m; ++i) c[i] = ;
         for (i=; i<=n; ++i) c[ x[y[i]] ]++;
         for (i=; i<=m; ++i) c[i] += c[i-];
         for (i=n; i>=; --i) sa[c[ x[y[i]] ]--] = y[i];
         swap(x,y);
         p = ;
         x[sa[]] = ;
         for (i=; i<=n; ++i)
             x[sa[i]] = y[sa[i]]==y[sa[i-]]&&y[sa[i]+k]==y[sa[i-]+k] ? p- : p++;
         if (p > n) break;
         m = p;
     }
 }
 void get_height() {
     for (int i=; i<=n; ++i) rnk[sa[i]] = i;
     int k = ;
     height[] = ;
     for (int i=; i<=n; ++i) {
         if (rnk[i]==) continue;
         if (k) k--;
         int j = sa[rnk[i]-];
         while (i+k<=n && j+k<=n && s[i+k]==s[j+k])
          k++;
         height[rnk[i]] = k;
     }
 }
 void get_ans() {
     long long ans = ;
     for (int i=; i<=n; ++i) {
         ans += max(n-i+-height[rnk[i]],);
 //      ans += max(n-sa[i]-height[i],0);
     }
     cout << ans << '\n';
 }
 int main() {
     int T;
     scanf("%d",&T);
     while (T--) {
         scanf("%s",s+);
         n = strlen(s+);
         get_sa();
         get_height();
         get_ans();
     }
     return ;
 }

原来的

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<cmath>

 using namespace std;

 const int N = ;
 char s[N];
 int t1[N],t2[N],sa[N],height[N],rnk[N],c[N];
 int n;

 void get_sa(int m) {
     int *x = t1,*y = t2;
     for (int i=; i<m; ++i) c[i] = ;
     for (int i=; i<n; ++i) c[x[i] = s[i]]++;
     for (int i=; i<m; ++i) c[i] += c[i-];
     for (int i=n-; i>=; --i) sa[--c[x[i]]] = i;
     for (int k=; k<=n; k<<=) {
         int p = ;
         for (int i=n-k; i<n; ++i) y[p++] = i;
         for (int i=; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;
         for (int i=; i<m; ++i) c[i] = ;
         for (int i=; i<n; ++i) c[x[y[i]]]++;
         for (int i=; i<m; ++i) c[i] += c[i-];
         for (int i=n-; i>=; --i) sa[--c[x[y[i]]]] = y[i];
         swap(x,y);
         p = ;
         x[sa[]] = ;
         for (int i=; i<n; ++i)
             x[sa[i]] = (y[sa[i-]]==y[sa[i]] && sa[i-]+k<n && sa[i]+k<n &&
             y[sa[i-]+k]==y[sa[i]+k]) ? p- : p++;

         if (p >= n) break;
         m = p;
     }
 }
 void get_height() {
     for (int i=; i<n; ++i) rnk[sa[i]] = i;
     int k = ;
     height[] = ;
     for (int i=; i<n; ++i) {
         if (!rnk[i]) continue;
         if (k) k--;
         int j = sa[rnk[i]-];
         while (i+k < n && j+k < n && s[i+k]==s[j+k]) k++;
         height[rnk[i]] = k;
     }
 }
 void get_ans() {
     long long ans = ;
     for (int i=; i<n; ++i) {
         ans += max(n-i-height[rnk[i]],);
 //        ans += max(n-sa[i]-height[i],0);
     }
     cout << ans << '\n';
 }
 int main() {
     int T;
     scanf("%d",&T);
     while (T--) {
         scanf("%s",s);
         n = strlen(s);
         get_sa();
         get_height();
         get_ans();
     }
     return ;
 }