SPOJ NSUBSTR (后缀自动机)

最新推荐文章于 2017-11-22 13:34:55 发布

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最新推荐文章于 2017-11-22 13:34:55 发布

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原文链接：http://www.cnblogs.com/rpSebastian/p/7218632.html

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SPOJ NSUBSTR

Problem : 给一个长度为n的字符串，要求分别输出长度为1~n的子串的最多出现次数。
Solution :首先对字符串建立后缀自动机，在根据fail指针建立出后缀树，对于n个后缀对应的节点打上标记，则每个结点对应的一些子串所出现的次数即为其子树内标记的个数，在后缀树上进行1遍dfs统计，之后根据答案的单调性线性扫描一遍进行统计。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

const int N = 250008;

struct node
{
    int u, v, nt;
};

struct Suffix_Automaton
{
    int nt[N << 1][26], a[N << 1], fail[N << 1];
    int tot, last, root;
    int p, q, np, nq;
    
    int lt[N << 1], sum;
    node eg[N << 2]; 
    int dp[N << 1], cnt[N << 1];

    void add(int u, int v)
    { 
        eg[++sum] = (node){u, v, lt[u]}; lt[u] = sum;
        eg[++sum] = (node){v, u, lt[v]}; lt[v] = sum;
    }
    int newnode(int len)
    {
        for (int i = 0; i < 26; ++i) nt[tot][i] = -1;
        fail[tot] = -1; a[tot] = len;
        return tot++;
    }
    void clear()
    {
        memset(lt, 0, sizeof(lt));
        tot = last = 0; sum = 0;
        root = newnode(0);
    }
    void insert(int ch)
    {
        p = last; last = np = newnode(a[p] + 1); cnt[np] = 1;
        for (; ~p && nt[p][ch] == -1; p = fail[p]) nt[p][ch] = np;
        if (p == -1) fail[np] = root;
        else
        {
            q = nt[p][ch];
            if (a[p] + 1 == a[q]) fail[np] = q;
            else
            {
                nq = newnode(a[p] + 1);
                for (int i = 0; i < 26; ++i) nt[nq][i] = nt[q][i];
                fail[nq] = fail[q];
                fail[np] = fail[q] = nq;
                for (; ~p && nt[p][ch] == q; p = fail[p]) nt[p][ch] = nq;
            }
        }
    }
    void dfs(int u, int fa)
    {
        for (int i = lt[u]; i; i = eg[i].nt)
        {
            int v = eg[i].v;
            if (v != fa)
            {
                dfs(v, u);
                cnt[u] += cnt[v];
            }
        }
        dp[a[u]] = max(dp[a[u]], cnt[u]);
    }
    void solve(int len)
    {
        for (int i = 1; i < tot; ++i) add(fail[i], i);
        dfs(root, -1);
        for (int i = len - 1; i >= 1; i--) dp[i] = max(dp[i], dp[i + 1]);
        for (int i = 1; i <= len; ++i) printf("%d\n", dp[i]);
    }
}sam;

int main()
{
    cin.sync_with_stdio(0);
    sam.clear();
    string s; cin >> s;
    for (int i = 0, len = s.length(); i < len; ++i)
        sam.insert(s[i] - 'a');
    sam.solve(s.length());
}

转载于:https://www.cnblogs.com/rpSebastian/p/7218632.html