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。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int sa[maxn],rk[maxn],height[maxn];
int c[maxn],t[maxn],t2[maxn],n;
char s[maxn];
void build_sa(int m) {
    int i,j,*x = t,*y = t2;
    for(i = 0; i < m; ++i) c[i] = 0;
    for(i = 0; i < n; ++i) c[x[i] = s[i]]++;
    for(i = 1; i < m; ++i) c[i] += c[i-1];
    for(i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;

    for(int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for(i = n-k; i < n; ++i) y[p++] = i;
        for(i = 0; i < n; ++i)
            if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; ++i) c[i] = 0;
        for(i = 0; i < n; ++i) c[x[y[i]]]++;
        for(i = 1; i < m; ++i) c[i] += c[i-1];
        for(i = n-1; i >= 0; --i)
            sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        x[sa[0]] = 0;
        p = 1;
        for(i = 1; i < n; ++i)
            if(y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k])
                x[sa[i]] = p-1;
            else x[sa[i]] = p++;
        if(p >= n) break;
        m = p;
    }
}
void getHeight(){
    int i,j,k = 0;
    for(i = 0; i < n; ++i) rk[sa[i]] = i;
    for(i = 0; i < n; ++i){
        if(k) --k;
        j = sa[rk[i]-1];
        while(i+k<n&&j+k<n&&s[i+k]==s[j+k]) ++k;
        height[rk[i]] = k;
    }
}
int main() {
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%s",s);
        n = strlen(s) + 1;
        build_sa(128);
        getHeight();
        int ret = 0;
        for(int i = 1; i < n; ++i)
            ret += n - sa[i] - height[i] - 1;
        printf("%d
",ret);
    }
    return 0;
}

后缀自动机

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2010;
char str[maxn];
int ret,n;
struct SAM{
    struct node{
        int son[26],f,len;
        void init(){
            f = -1;
            len = 0;
            memset(son,-1,sizeof son);
        }
    }e[maxn];
    int tot,last;
    int newnode(int len = 0){
        e[tot].init();
        e[tot].len = len;
        return tot++;
    }
    void init(){
        tot = last = 0;
        newnode();
    }
    void extend(int c){
        int p = last,np = newnode(e[p].len + 1);
        while(p != -1 && e[p].son[c] == -1){
            e[p].son[c] = np;
            p = e[p].f;
        }
        if(p == -1) e[np].f = 0;
        else{
            int q = e[p].son[c];
            if(e[p].len + 1 == e[q].len) e[np].f = q;
            else{
                int nq = newnode();
                e[nq] = e[q];
                e[nq].len = e[p].len + 1;
                e[q].f = e[np].f = nq;
                while(p != -1 && e[p].son[c] == q){
                    e[p].son[c] = nq;
                    p = e[p].f;
                }
            }
        }
        last = np;
    }
    void solve(){
        ret = 0;
        for(int i = 1; i < tot; ++i)
            ret += e[i].len - e[e[i].f].len;
        printf("%d
",ret);
    }
}sam;

int main(){
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%s",str);
        n = strlen(str);
        sam.init();
        for(int i = ret = 0; str[i]; ++i)
            sam.extend(str[i]-'A');
        sam.solve();
    }
    return 0;
}