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Given a string, we need to find the total number of its distinct substrings.
T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000
For each test case output one number saying the number of distinct substrings.
Input: 2 CCCCC ABABA Output: 5 9
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 long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 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 }
SPOJ 705 New Distinct Substrings
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原文地址:http://www.cnblogs.com/crackpotisback/p/4033744.html