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凤兮凤兮归故乡 遨游四海求其凰

题目链接

  GSAM:求N个串的本质不同的子串的数目。

  于是,如果单串的SAM肯定就不够用了,所以我们不妨利用构造一棵字典树,然后在字典树上利用BFS去跑SAM来实现GSAM。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define pii pair<int, int>
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxS = 1e6 + 7;
int N, tot, root;
struct Trie
{
    int c, fa;
    int nex[26];
} t[maxS];
char s[maxS];
void Insert()
{
    int len = (int)strlen(s);
    int rt = root;
    for(int i = 0, id; i < len; i++)
    {
        id = s[i] - 'a';
        if(!t[rt].nex[id])
        {
            t[rt].nex[id] = ++tot;
            t[tot].fa = rt;
            t[tot].c = id;
        }
        rt = t[rt].nex[id];
    }
}
struct SAM
{
    struct state
    {
        int len, link, next[30];
    } st[maxS << 1];
    int siz = 1;
    void init()
    {
        siz = 1;
        st[1].len = 0;
        st[1].link = 0;
        siz++;
    }
    int extend(int c, int last)
    {
        int cur = siz++;
        st[cur].len = st[last].len + 1;
        int p = last;
        while (p != 0 && !st[p].next[c])
        {
            st[p].next[c] = cur;
            p = st[p].link;
        }
        if (p == 0)
        {
            st[cur].link = 1;
        }
        else
        {
            int q = st[p].next[c];
            if (st[p].len + 1 == st[q].len)
            {
                st[cur].link = q;
            }
            else
            {
                int clone = siz++;
                st[clone].len = st[p].len + 1;
                memcpy(st[clone].next, st[q].next, sizeof(st[q].next));
                st[clone].link = st[q].link;
                while (p != 0 && st[p].next[c] == q)
                {
                    st[p].next[c] = clone;
                    p = st[p].link;
                }
                st[q].link = st[cur].link = clone;
            }
        }
        return last = cur;
    }
} sam;
int pos[maxS];
int main()
{
    scanf("%d", &N);
    tot = 0;
    root = ++tot;
    for(int i=1; i<=N; i++)
    {
        scanf("%s", s);
        Insert();
    }
    sam.init();
    queue<int> Q;
    for(int i=0; i<26; i++) if(t[root].nex[i]) Q.push(t[root].nex[i]);
    pos[root] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        pos[u] = sam.extend(t[u].c, pos[t[u].fa]);
        for(int i=0; i<26; i++) if(t[u].nex[i]) Q.push(t[u].nex[i]);
    }
    ll ans = 0;
    for(int i=2; i<=sam.siz; i++) ans += sam.st[i].len - sam.st[sam.st[i].link].len;
    printf("%lld\n", ans);
    return 0;
}

 

posted on 2020-09-19 10:29  唔哩Wulili  阅读(158)  评论(0编辑  收藏  举报