Spoj 705 new distinct substrings

New distinct substringstime limit: 2000 msmemory limit: 262144 kbthis problem will be judged on spoj. Original ID: subst1
64-bit integer Io format: % LLD Java class name: Main

Given a string, we need to find the total number of its distinct substrings.

Input

T-Number of test cases. T <= 20; each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:2CCCCCABABAOutput:59

Sourcebase on a problem in bytecode06: calculate the number of different substrings. The application of the suffix array LCP.

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 50010;18 int rk[maxn],wb[maxn],wv[maxn],wd[maxn],lcp[maxn];19 bool cmp(int *r,int i,int j,int k){20     return r[i] == r[j] && r[i+k] == r[j+k];21 }22 void da(int *r,int *sa,int n,int m){23     int i,k,p,*x = rk,*y = wb;24     for(i = 0; i < m; ++i) wd[i] = 0;25     for(i = 0; i < n; ++i) wd[x[i] = r[i]]++;26     for(i = 1; i < m; ++i) wd[i] += wd[i-1];27     for(i = n-1; i >= 0; --i) sa[--wd[x[i]]] = i;28 29     for(p = k = 1; p < n; k <<= 1,m = p){30         for(p = 0,i = n-k; i < n; ++i) y[p++] = i;31         for(i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;32         for(i = 0; i < n; ++i) wv[i] = x[y[i]];33 34         for(i = 0; i < m; ++i) wd[i] = 0;35         for(i = 0; i < n; ++i) wd[wv[i]]++;36         for(i = 1; i < m; ++i) wd[i] += wd[i-1];37         for(i = n-1; i >= 0; --i) sa[--wd[wv[i]]] = y[i];38 39         swap(x,y);40         x[sa[0]] = 0;41         for(p = i = 1; i < n; ++i)42             x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;43     }44 }45 void calcp(int *r,int *sa,int n){46     for(int i = 1; i <= n; ++i) rk[sa[i]] = i;47     int h = 0;48     for(int i = 0; i < n; ++i){49         if(h > 0) h--;50         for(int j = sa[rk[i]-1]; i+h < n && j+h < n; h++)51             if(r[i+h] != r[j+h]) break;52         lcp[rk[i]-1] = h;53     }54 }55 int r[maxn],sa[maxn];56 char str[maxn];57 int main() {58     int n;59     scanf("%d",&n);60     while(n--){61         scanf("%s",str);62         int len = strlen(str);63         for(int i = 0; i < len; ++i)64             r[i] = str[i];65         r[len] = 0;66         da(r,sa,len+1,128);67         calcp(r,sa,len);68         int ans = 0;69         for(int i = 1; i <= len; ++i)70             ans += len - sa[i] - lcp[i];71         printf("%d\n",ans);72     }73     return 0;74 }

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Spoj 705 new distinct substrings