• SPOJ 694 Distinct Substrings


    Distinct Substrings

    Time Limit: 1000ms
    Memory Limit: 262144KB
    This problem will be judged on SPOJ. Original ID: DISUBSTR
    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 <= 1000

    Output

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

    Example

    Sample Input:
    2
    CCCCC
    ABABA

    Sample Output:
    5
    9

    Explanation for the testcase with string ABABA: 
    len=1 : A,B
    len=2 : AB,BA
    len=3 : ABA,BAB
    len=4 : ABAB,BABA
    len=5 : ABABA
    Thus, total number of distinct substrings is 9.

    解题:求不同子串的个数,n - sa[i] - 1表示当前后缀的前缀个数,然后减去 height[i],也就是减去与前面的重复的。

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 100010;
     4 int sa[maxn],rk[maxn],height[maxn];
     5 int c[maxn],t[maxn],t2[maxn],n;
     6 char s[maxn];
     7 void build_sa(int m) {
     8     int i,j,*x = t,*y = t2;
     9     for(i = 0; i < m; ++i) c[i] = 0;
    10     for(i = 0; i < n; ++i) c[x[i] = s[i]]++;
    11     for(i = 1; i < m; ++i) c[i] += c[i-1];
    12     for(i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;
    13 
    14     for(int k = 1; k <= n; k <<= 1) {
    15         int p = 0;
    16         for(i = n-k; i < n; ++i) y[p++] = i;
    17         for(i = 0; i < n; ++i)
    18             if(sa[i] >= k) y[p++] = sa[i] - k;
    19         for(i = 0; i < m; ++i) c[i] = 0;
    20         for(i = 0; i < n; ++i) c[x[y[i]]]++;
    21         for(i = 1; i < m; ++i) c[i] += c[i-1];
    22         for(i = n-1; i >= 0; --i)
    23             sa[--c[x[y[i]]]] = y[i];
    24         swap(x,y);
    25         x[sa[0]] = 0;
    26         p = 1;
    27         for(i = 1; i < n; ++i)
    28             if(y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k])
    29                 x[sa[i]] = p-1;
    30             else x[sa[i]] = p++;
    31         if(p >= n) break;
    32         m = p;
    33     }
    34 }
    35 void getHeight(){
    36     int i,j,k = 0;
    37     for(i = 0; i < n; ++i) rk[sa[i]] = i;
    38     for(i = 0; i < n; ++i){
    39         if(k) --k;
    40         j = sa[rk[i]-1];
    41         while(i+k<n&&j+k<n&&s[i+k]==s[j+k]) ++k;
    42         height[rk[i]] = k;
    43     }
    44 }
    45 int main() {
    46     int kase;
    47     scanf("%d",&kase);
    48     while(kase--){
    49         scanf("%s",s);
    50         n = strlen(s) + 1;
    51         build_sa(128);
    52         getHeight();
    53         int ret = 0;
    54         for(int i = 1; i < n; ++i)
    55             ret += n - sa[i] - height[i] - 1;
    56         printf("%d
    ",ret);
    57     }
    58     return 0;
    59 }
    View Code

    后缀自动机

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 2010;
     4 char str[maxn];
     5 int ret,n;
     6 struct SAM{
     7     struct node{
     8         int son[26],f,len;
     9         void init(){
    10             f = -1;
    11             len = 0;
    12             memset(son,-1,sizeof son);
    13         }
    14     }e[maxn];
    15     int tot,last;
    16     int newnode(int len = 0){
    17         e[tot].init();
    18         e[tot].len = len;
    19         return tot++;
    20     }
    21     void init(){
    22         tot = last = 0;
    23         newnode();
    24     }
    25     void extend(int c){
    26         int p = last,np = newnode(e[p].len + 1);
    27         while(p != -1 && e[p].son[c] == -1){
    28             e[p].son[c] = np;
    29             p = e[p].f;
    30         }
    31         if(p == -1) e[np].f = 0;
    32         else{
    33             int q = e[p].son[c];
    34             if(e[p].len + 1 == e[q].len) e[np].f = q;
    35             else{
    36                 int nq = newnode();
    37                 e[nq] = e[q];
    38                 e[nq].len = e[p].len + 1;
    39                 e[q].f = e[np].f = nq;
    40                 while(p != -1 && e[p].son[c] == q){
    41                     e[p].son[c] = nq;
    42                     p = e[p].f;
    43                 }
    44             }
    45         }
    46         last = np;
    47     }
    48     void solve(){
    49         ret = 0;
    50         for(int i = 1; i < tot; ++i)
    51             ret += e[i].len - e[e[i].f].len;
    52         printf("%d
    ",ret);
    53     }
    54 }sam;
    55 
    56 int main(){
    57     int kase;
    58     scanf("%d",&kase);
    59     while(kase--){
    60         scanf("%s",str);
    61         n = strlen(str);
    62         sam.init();
    63         for(int i = ret = 0; str[i]; ++i)
    64             sam.extend(str[i]-'A');
    65         sam.solve();
    66     }
    67     return 0;
    68 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4634251.html
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