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Chaos Slover 补档计划 - 后缀自动机（2）

2015.11.17 09:42 Tue| 23 visits oi_2016| ChaosSlover补档计划| Text

据说都是后缀数组的经典题

POJ1743 Musical Theme

可以使用后缀数组+二分答案。

也可以使用后缀自动机，处理出每一个结点的 right 集合中的最大和最小值，它们的差-1（其实不 -1 也能过，可是这里真的需要 -1）和长度之间的最小值就是这个结点上的答案。

注意最后答案需要 +1，因为在最开始差分了。

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#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;

int n, a[20005];

struct node
{
    node *son[175], *pre;
    int len, minr, maxr;
    node(int _len = 0) : len(_len), minr(20000), maxr(0)
    {
        memset(son, 0, sizeof son); pre = 0;
    }
    void *operator new(size_t s);
} newnode[40001], *pos;

void *node::operator new(size_t s)
{
    return pos++;
}

node *root, *last;
vector<node *> v[20005];

void reset()
{
    for (int i = 0; i <= n; ++i)
        v[i].resize(0);
    pos = newnode;
    last = root = new node();
}

void insert(int c)
{
    node *p = last, *np = new node(p->len + 1);
    while (p && !p->son[c])
        p->son[c] = np, p = p->pre;
    if (!p) np->pre = root;
    else
    {
        node *q = p->son[c];
        if (q->len == p->len + 1) np->pre = q;
        else
        {
            node *nq = new node(p->len + 1);
            memcpy(nq->son, q->son, sizeof nq->son);
            nq->pre = q->pre;
            np->pre = q->pre = nq;
            while (p && p->son[c] == q)
                p->son[c] = nq, p = p->pre;
        }
    }
    last = np;
}

int getright()
{
    int re = 0;
    node *p = root;
    p->maxr = p->minr = 0;
    for (int i = 0; i < n; ++i)
        p = p->son[a[i]], p->maxr = p->minr = i + 1;
    for (node *i = newnode; i != pos; ++i)
        v[i->len].push_back(i);
    for (int i = n; i >= 1; --i)
        for (int j = 0; j < (int)v[i].size(); ++j)
            re = max(re, min(v[i][j]->len, v[i][j]->maxr - v[i][j]->minr - 1)),
            v[i][j]->pre->minr = min(v[i][j]->pre->minr, v[i][j]->minr),
            v[i][j]->pre->maxr = max(v[i][j]->pre->maxr, v[i][j]->maxr);
    return re < 4 ? 0 : re + 1;
}

int main()
{
    while (cin >> n, n)
    {
        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]); --n;
        for (int i = 0; i < n; ++i)
            a[i] = a[i + 1] - a[i] + 87;
        reset();
        for (int i = 0; i < n; ++i)
            insert(a[i]);
        cout << getright() << endl;
    }
}

SPOJSUBST1+DISUBSTR New Distinct Substrings

题目里面并没有说只有大写字母，真是令人悲伤的事情啊！

问互不相同的子串个数，其实就是每一个结点能够表示的字符串个数的和。p->len-p->pre->len 加起来就好了。

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#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

struct node
{
    node *son[128], *pre;
    int len;
    node(int _len = 0) : len(_len)
    {
        memset(son, 0, sizeof son); pre = 0;
    }
    void *operator new(size_t s);
} newnode[100010], *pos;

void *node::operator new(size_t s)
{
    return pos++;
}

node *root, *last;
int n, ans; char str[50010];

void reset()
{
    pos = newnode; ans = 0;
    last = root = new node();
}

void insert(int c)
{
    node *p = last, *np = new node(p->len + 1);
    while (p && !p->son[c])
        p->son[c] = np, p = p->pre;
    if (!p) np->pre = root;
    else
    {
        node *q = p->son[c];
        if (q->len == p->len + 1) np->pre = q;
        else
        {
            node *nq = new node(p->len + 1);
            memcpy(nq->son, q->son, sizeof nq->son);
            nq->pre = q->pre;
            np->pre = q->pre = nq;
            while (p && p->son[c] == q)
                p->son[c] = nq, p = p->pre;
        }
    }
    last = np; ans += np->len - np->pre->len;
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        scanf("%s", str);
        reset();
        int l = strlen(str);
        for (int i = 0; i < l; ++i)
            insert(str[i]);
        cout << ans << endl;
    }
    return 0;
}

BZOJ3676 回文串

用 Manacher 求出来所有的本质不同的回文串，然后放进后缀自动机里面跑。

本质不同的回文串实际上就是使得 maxp 增大的回文串，很显然最多也只有 N 个。注意使用 manacher 的时候好好判断边界 x_x 并且只能统计边缘是字母的回文串。

然后还需要在后缀自动机里面输出匹配的结点的 right 大小。这里可以用倍增迅速找到这个结点。预处理出从每一个结点开始沿着 pre 边向上跳 2^k 次能到达的位置，然后通过 len 的大小判断是否能跳。

然而这道题是一道回文自动机的裸题！

然而这道题卡后缀数组的时限！

然而这道题卡后缀自动机的内存！

然而这道题还需要开 long long！

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#include <bits/stdc++.h>
using namespace std;

struct node
{
    node *son[26], *pre;
    int len, right, id;
    node(int _len = 0) : len(_len) {}
    void *operator new(size_t s);
} newnode[600005], *pos;

void *node::operator new(size_t s)
{
    return pos++;
}

node *root, *last, *top[600005];
vector<node *> v[300005];
int n, tot, fa[600005][18], p[600005], point[300005];
char str[300005], s[600005];

void reset()
{
    pos = newnode;
    last = root = new node();
}

void insert(int c)
{
    node *p = last, *np = new node(p->len + 1);
    while (p && !p->son[c])
        p->son[c] = np, p = p->pre;
    if (!p) np->pre = root;
    else
    {
        node *q = p->son[c];
        if (q->len == p->len + 1) np->pre = q;
        else
        {
            node *nq = new node(p->len + 1);
            memcpy(nq->son, q->son, sizeof nq->son);
            nq->pre = q->pre;
            np->pre = q->pre = nq;
            while (p && p->son[c] == q)
                p->son[c] = nq, p = p->pre;
        }
    }
    last = np;
}

void build()
{
    for (node *i = newnode; i != pos; ++i)
        v[i->len].push_back(i);
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j < (int)v[i].size(); ++j)
            top[tot] = v[i][j], v[i][j]->id = tot++;
    for (int i = 1; i < tot; ++i)
        fa[i][0] = top[i]->pre->id;
    for (int j = 1; j <= 17; ++j)
        for (int i = 1; i < tot; ++i)
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
    node *p = root;
    for (int i = 0; i < n; ++i)
        p = p->son[str[i] - 'a'], p->right = 1,
        point[i + 1] = p->id;
    for (int i = tot - 1; i; --i)
        top[i]->pre->right += top[i]->right;
}

long long query(int l, int r)
{
    int len = r - l + 1;
    int x = point[r];
    for (int i = 17; i >= 0; --i)
        if (top[fa[x][i]]->len >= len)
            x = fa[x][i];
    return 1ll * top[x]->right * len;
}

long long manacher()
{
    s[0] = '$'; s[1] = '#';
    for (int i = 0; i < n; ++i)
        s[i * 2 + 2] = str[i], s[i * 2 + 3] = '#';
    int len = n * 2 + 2, maxp = 0, id = 0;
    long long re = 0;
    for (int i = 1; i < len; ++i)
    {
        if (maxp > i) p[i] = min(p[id * 2 - i], maxp - i);
        else
        {
            p[i] = 1;
            if (s[i] != '#')
                re = max(re, query((i - p[i] + 2) / 2, (i + p[i]) / 2));
        }
        while (s[i - p[i]] == s[i + p[i]])
        {
            ++p[i];
            if (s[i + p[i] - 1] != '#')
                re = max(re, query((i - p[i] + 2) / 2, (i + p[i]) / 2));
        }
        if (i + p[i] > maxp) maxp = i + p[i], id = i;
    }
    return re;
}

int main()
{
    scanf("%s", str);
    n = strlen(str);
    reset();
    for (int i = 0; i < n; ++i)
        insert(str[i] - 'a');
    build();
    cout << manacher() << endl;
    return 0;
}

POJ3294 Life Forms（伪）

广义后缀自动机，然而本人并不会输出方案，更不知道这份代码的正确性。这道题还是该用后缀数组做吧。

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#include <bits/stdc++.h>
using namespace std;

struct node
{
    node *son[26], *pre;
    int len;
    bitset<105> vis;
    node(int _len = 0) : len(_len)
    {
        vis.reset();
        memset(son, 0, sizeof son); pre = 0;
    }
    void *operator new(size_t s);
} newnode[200005], *pos;

void *node::operator new(size_t s)
{
    return pos++;
}

int n; char str[1005];
vector<node *> v[1005];
node *root, *last;

void reset()
{
    pos = newnode;
    last = root = new node();
}

void insert(int id, int c)
{
    node *p = last;
    if (p->son[c])
    {
        node *q = p->son[c];
        if (q->len == p->len + 1) last = q;
        else
        {
            node *nq = new node(p->len + 1);
            memcpy(nq->son, q->son, sizeof nq->son);
            nq->pre = q->pre; q->pre = nq;
            while (p && p->son[c] == q)
                p->son[c] = nq, p = p->pre;
            last = nq;
        }
    }
    else
    {
        node *np = new node(p->len + 1);
        while (p && !p->son[c])
            p->son[c] = np, p = p->pre;
        if (!p) np->pre = root;
        else
        {
            node *q = p->son[c];
            if (q->len == p->len + 1) np->pre = q;
            else
            {
                node *nq = new node(p->len + 1);
                memcpy(nq->son, q->son, sizeof nq->son);
                nq->pre = q->pre;
                np->pre = q->pre = nq;
                while (p && p->son[c] == q)
                    p->son[c] = nq, p = p->pre;
            }
        }
        last = np;
    }
    last->vis[id] = 1;
}

int main()
{
    while (cin >> n, n)
    {
        reset();
        for (int i = 1; i <= n; ++i)
        {
            last = root;
            scanf("%s", str);
            int l = strlen(str);
            for (int j = 0; j < l; ++j)
                insert(i, str[j] - 'a');
        }
        int maxl = 0;
        for (int i = 0; i <= 1000; ++i)
            v[i].resize();
        for (node *i = newnode; i != pos; ++i)
            v[i->len].push_back(i);
        for (int i = 1000; i; --i)
            for (auto j : v[i])
                j->pre->vis |= j->vis;
        int ans = 0;
        for (node *i = newnode; i != pos; ++i)
            if ((int)i->vis.count() > (n >> 1))
                ans = max(ans, i->len);
        if (ans == 0) puts("?");
        else cout << ans << endl;
    }
}

Update: POJ3294 Life Forms

用后缀数组二分答案将 height 分组。细节就是需要分配不同的分隔符。另外这题用 C++ 交蜜汁 RE，G++ 通过。

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#include <bits/stdc++.h>
using namespace std;

#define N 110005

char str[N];
int x[N], y[N], c[N], tot, vis[105];
int s, n, bel[N], sa[N], rnk[N], height[N];

inline bool same(int a, int b, int l)
{
    return y[a] == y[b] &&
                ((a + l >= n && b + l >= n) ||
                (a + l < n && b + l < n && y[a + l] == y[b + l]));
}

void get_sa(int m = 256)
{
    for (int i = 0; i < n; ++i)         ++c[x[i] = 128 + str[i]];
    for (int i = 1; i < m; ++i)         c[i] += c[i - 1];
    for (int i = n - 1; i >= 0; --i)    sa[--c[x[i]]] = i;
    for (int k = 1; k < n; k <<= 1)
    {
        int p = 0;
        for (int i = n - k; i < n; ++i)     y[p++] = i;
        for (int i = 0; i < n; ++i)         if (sa[i] >= k) y[p++] = sa[i] - k;
        memset(c, 0, m * sizeof(int));
        for (int i = 0; i < n; ++i)         ++c[x[y[i]]];
        for (int i = 1; i < m; ++i)         c[i] += c[i - 1];
        for (int i = n - 1; i >= 0; --i)    sa[--c[x[y[i]]]] = y[i];
        for (int i = 0; i < n; ++i)         y[i] = x[i];
        m = 0; x[sa[0]] = 0;
        for (int i = 1; i < n; ++i)
            x[sa[i]] = same(sa[i], sa[i - 1], k) ? m : ++m;
        if (++m == n) break;
    }
}

void get_height()
{
    for (int i = 0; i < n; ++i) rnk[sa[i]] = i;
    for (int i = 0, j, k = 0; i < n; height[rnk[i++]] = k)
        if (rnk[i])
        {
            k = max(k - 1, 0); j = sa[rnk[i] - 1];
            while (str[i + k] == str[j + k]) ++k;
        }
}

void init()
{
    memset(sa, 0, sizeof sa);
    memset(rnk, 0, sizeof rnk);
    memset(height, 0, sizeof height);
    memset(x, 0, sizeof x);
    memset(y, 0, sizeof y);
    memset(c, 0, sizeof c);
}

bool check(int mid)
{
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
    {
        if (height[i] < mid)
        {
            if (cnt > s / 2) return true;
            cnt = 0; ++tot;
        }
        else
        {
            if (vis[bel[sa[i - 1]]] != tot)
                ++cnt, vis[bel[sa[i - 1]]] = tot;
            if (vis[bel[sa[i]]] != tot)
                ++cnt, vis[bel[sa[i]]] = tot;
        }
    }
    return false;
}

void print(int ans)
{
    int cnt = 0;
    for (int i = 1; i <= n; ++i)
    {
        if (height[i] < ans)
        {
            if (cnt > s / 2)
            {
                for (int j = 0; j < ans; ++j)
                    putchar(str[sa[i - 1] + j]);
                puts("");
            }
            cnt = 0; ++tot;
        }
        else
        {
            if (vis[bel[sa[i - 1]]] != tot)
                ++cnt, vis[bel[sa[i - 1]]] = tot;
            if (vis[bel[sa[i]]] != tot)
                ++cnt, vis[bel[sa[i]]] = tot;
        }
    }
}

int main()
{
    while (cin >> s, s)
    {
        init();
        n = -1;
        for (int i = 1; i <= s; ++i)
        {
            str[n] = 150 + i,
            scanf("%s", str + n + 1);
            int t = n;
            n = strlen(str);
            for (int j = t + 1; j <= n; ++j)
                bel[j] = i;
        }
        get_sa();
        get_height();
        int l = 1, r = n, ans = 0;
        while (l <= r)
        {
            int mid = (l + r) >> 1;
            if (check(mid)) l = mid + 1, ans = mid;
            else r = mid - 1;
        }
        if (ans == 0) puts("?");
        else print(ans);
        puts("");
    }
    return 0;
}
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