后缀自动机合集 I

简介

有关后缀自动机的理论和证明各路大神的博客多如牛毛，就不细说。

贴图太占载入速度了，还是去看原版的解析吧  http://hi.baidu.com/myidea/item/142c5cd45901a51820e25039

对于每个State点，father指向与该节点可接收相同后缀的节点，其表示的字串通常是当前节点表示字串的后面部分，

因此State->step - Sate->father->step 代表从任一后缀串起始走到该状态的路径数

hdu 4622 Reincarnation

对每个状态点减去父状态的步骤数的和就是字符串所有字串的数量

#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 2010;int ans[MAXN][MAXN];char s[MAXN];struct node{    node*nxt[26], *fa;    int step;    void clear()    {        fa = 0, step = 0;        memset ( nxt, 0, sizeof nxt );    }    int calc()    {        if ( !fa ) { return 0; }        return step - fa->step;    }} *root, *tail, nodePool[MAXN * 2], *cur;void init(){    cur = nodePool;    root = tail = cur++;    root->clear();}int tot;void Insert ( int w ){    node *p = tail, *np = cur++;    np->clear();    np->step = p->step + 1;    for ( ; p && !p->nxt[w]; p = p->fa ) { p->nxt[w] = np; }    if ( !p )    {        np->fa = root;        tot += np->calc();    }    else    {        if ( p->nxt[w]->step == p->step + 1 )        {            np->fa = p->nxt[w];            tot += np->calc();        }        else        {            node *q = p->nxt[w], *r = cur++;            *r = *q;            tot -= q->calc();            r->step = p->step + 1;            q->fa = np->fa = r;            tot += q->calc() + r->calc() + np->calc();            for ( ; p && p->nxt[w] == q; p = p->fa ) { p->nxt[w] = r; }        }    }    tail = np;}int main(){#ifdef  __GNUC__    freopen ( "in.txt", "r", stdin );#endif // __GNUC__    int t;    scanf ( "%d", &t );    while ( t-- )    {        scanf ( "%s", s );        int n = strlen ( s );        for ( int i = 0; i < n; ++i )        {            init();            tot = 0;            for ( int j = i; j < n; ++j )            {                Insert ( s[j] - 'a' );                ans[i][j] = tot;            }        }                int q;        scanf ( "%d", &q );        while ( q-- )        {            int l, r;            scanf ( "%d%d", &l, &r );            l--, r--;            printf ( "%d\n", ans[l][r] );        }    }    return 0;}

hdu 1403 Longest Common Substring

每个step姑且认为是转移到该状态时可接收的最大字符串

最长公共子串

#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 100010;char s[MAXN];struct node{    node*nxt[26], *fa;    int step;    void clear()    {        fa = 0, step = 0;        memset ( nxt, 0, sizeof nxt );    }} *root, *tail, nodePool[MAXN << 1], *cur;void init(){    cur = nodePool;    root = tail = cur++;    root->clear();}void Insert ( int w ){    node *p = tail, *np = cur++;    np->clear();    np->step = p->step + 1;    for ( ; p && !p->nxt[w]; p = p->fa ) { p->nxt[w] = np; }    if ( !p )    {        np->fa = root;    }    else    {        if ( p->nxt[w]->step == p->step + 1 )        {            np->fa = p->nxt[w];        }        else        {            node *q = p->nxt[w], *r = cur++;            *r = *q;            r->step = p->step + 1;            q->fa = np->fa = r;            for ( ; p && p->nxt[w] == q; p = p->fa ) { p->nxt[w] = r; }        }    }    tail = np;}int solve(){int l = strlen(s), w, res = 0, sum= 0;node *p = root;for (int i = 0; i< l; ++i){w = s[i]-'a';if ( p->nxt[w]){++sum;p = p->nxt[w];}else{while ( p && p->nxt[w] == NULL) p=p->fa;if ( p== NULL){sum = 0;p = root;}else{sum = p->step + 1;p = p->nxt[w];}}res = max(sum, res);}return res;}int main(){#ifdef  __GNUC__    freopen ( "in.txt", "r", stdin );#endif // __GNUC__    while ( scanf("%s", s) != EOF)    {        init();        int l = strlen(s);        for (int i = 0; i< l; ++i){Insert(s[i]-'a'); }scanf("%s", s);printf("%d\n", solve());    }    return 0;}

hdu 4416 Good Article Good sentence

菜鸟对SAM的理解还是太弱了，借鉴了别人的代码才搞懂

本题用SAM来做，核心是记录下到达每个State时能得到Bi字串的最大长度。

再此还要解释下，引入father指针的目的是最大化状态的重用性，每个状态的父节点表示的串 == 当前状态所表示的字符串最后几位

因此，父节点的deep要更新为自身及子节点deep的最大值

每个节点deep - step 就表示为除去不符合要求串数量的后的值

#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>using namespace std;typedef long long LL;const int MAXN = 100010;char s[MAXN];struct node{    node*nxt[26], *fa;    int step;    int deep;    void clear()    {        fa = 0, step = 0; deep = 0;        memset ( nxt, 0, sizeof nxt );    }} *root, *tail, nodePool[MAXN << 1], *cur;void init(){    cur = nodePool;    root = tail = cur++;    root->clear();}void Insert ( int w ){    node *p = tail, *np = cur++;    np->clear();    np->step = p->step + 1;    for ( ; p && !p->nxt[w]; p = p->fa ) { p->nxt[w] = np; }    if ( !p )    {        np->fa = root;    }    else    {        if ( p->nxt[w]->step == p->step + 1 )        {            np->fa = p->nxt[w];        }        else        {            node *q = p->nxt[w], *r = cur++;            *r = *q;            r->step = p->step + 1;            q->fa = np->fa = r;            for ( ; p && p->nxt[w] == q; p = p->fa ) { p->nxt[w] = r; }        }    }    tail = np;}void update(){    int l = strlen ( s ), w, sum = 0;    node *p = root;    for ( int i = 0; i < l; ++i )    {        w = s[i] - 'a';        if ( p->nxt[w] )        {            ++sum;            p = p->nxt[w];        }        else        {            while ( p && p->nxt[w] == NULL ) { p = p->fa; }            if ( p == NULL )            {                sum = 0;                p = root;            }            else            {                sum = p->step + 1;                p = p->nxt[w];            }        }        p->deep = max ( p->deep, sum );    }}int p[MAXN << 1]; node *op[MAXN << 1];int main(){#ifdef  __GNUC__    freopen ( "in.txt", "r", stdin );#endif // __GNUC__    int t, cs = 0;    int n;    scanf ( "%d", &t );        while ( t-- )    {        printf ( "Case %d: ", ++cs );        scanf ( "%d", &n );        scanf ( "%s", s );        init();        int l = strlen ( s );        for ( int i = 0; i < l; ++i ) { Insert ( s[i] - 'a' ); }        for ( int i = 0; i < n; ++i )        {            scanf ( "%s", s );            update();        }        LL res = 0;        memset ( p, 0, sizeof p );        for ( int i = 0; i < cur - root; ++i ) { p[nodePool[i].step]++; }        for ( int i = 1; i <= tail->step; ++i ) { p[i] += p[i - 1]; }        for ( int i = 0; i < cur - root; ++i ) { op[--p[nodePool[i].step]] = nodePool + i; }        for ( int i = cur - root - 1; i > 0; --i )        {            node *u = op[i];            if ( u->deep > 0 )            {                u->fa->deep = max ( u->fa->deep, u->deep );                if ( u->deep < u->step )                {                    res += u->step - u->deep ;                }            }            else            {                res += u->step - u->fa->step;            }        }#ifdef __GNUC__        printf ( "%lld\n", res );#else        printf ( "%I64d\n", res );#endif    }    return 0;}

hdu 4270 Dynamic Lover

关键问题是后缀自动机状态删除，可行的办法就是设置标记表示该状态是否被删除，且同一状态的拷贝点也指向相同的标记

#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 200020;char s[MAXN ];int isDel[MAXN], delCnt;struct node{    node*nxt[26], *fa;    int step;    int *isdel;    int pos;    node *flg;    void clear()    {        fa = 0, step = 0; isdel = 0; flg = 0;        memset ( nxt, 0, sizeof nxt );    }} *root, *tail, nodePool[MAXN << 1], *cur, *mseq[MAXN], *cnt;void init(){    memset ( isDel, 0, sizeof isDel );        cur = nodePool;    root = tail = cur++;    root->clear();    mseq[0] = root;    root->isdel = isDel;    delCnt = 1;}int judge ( node *x ){    return x == NULL || *x->isdel;}void Insert ( int w ){    node *p = tail, *np = cur++;    np->clear();    np->step = p->step + 1;    np->pos = np->step;    mseq[np->pos] = np;    np->isdel = isDel + delCnt++;    for ( ; p && judge ( p->nxt[w] ); p = p->fa ) { p->nxt[w] = np; }    if ( !p )    {        np->fa = root;    }    else    {        if ( p->nxt[w]->step == p->step + 1 )        {            np->fa = p->nxt[w];        }        else        {            node *q = p->nxt[w], *r = cur++;            *r = *q;            r->step = p->step + 1;            q->fa = np->fa = r;            for ( ; p && p->nxt[w] == q; p = p->fa ) { p->nxt[w] = r; }        }    }    tail = np;}int flag, tl;void dfs ( node *k, int l ){    if ( flag ) { return; }    if ( l == tl ) {flag = k->pos - l + 1; return; }    if ( k->flg == cnt )    {        flag = cnt->pos - l + 1;        return;    }    for ( int i = 0; !flag && i < 26; ++i )    {        if ( k->nxt[i] && !*k->nxt[i]->isdel )        {            dfs ( k->nxt[i], l + 1 );        }    }}int main(){#ifdef  __GNUC__    freopen ( "in.txt", "r", stdin );#endif // __GNUC__    int n, c, k;    char *p;    while ( scanf ( "%s", s ) != EOF )    {        init();        for ( p = s; *p; ++p ) { Insert ( *p - 'a' ); }        scanf ( "%d", &n );        while ( n-- )        {            scanf ( "%d", &c );            if ( c == 1 )            {                scanf ( "%s", s );                char *p = s;                for ( ; *p; ++p ) { Insert ( *p - 'a' ); }            }            else if ( c == 2 )            {                cnt = tail;                for ( node *as = tail; as != root ; as = as->fa )                {                    as->flg = cnt;                }                scanf ( "%d", &k );                flag = 0; tl = k;                dfs ( root, 0 );                printf ( "%d\n", flag );            }            else            {                scanf ( "%d", &k );                while ( k-- )                {                    *tail->isdel = 1;                    tail = mseq[tail->step - 1];                }            }        }    }    return 0;}