后缀自动机学习笔记

诸神眷顾的幻想乡

给定一个叶节点不超过20的无根树，每个节点有一个字母。问树上路径形成的本质不同的字符串的个数。

广义后缀自动机裸题。从每个叶节点做bfs，记录父亲的状态从而插入建立后缀自动机。我们知道一个后缀自动机本质不同的子串个数为 ∑imax[i]−min[i]+1=max[i]−max[fa[i]]，或者DAG上dp即可。

#include <bits/stdc++.h>using namespace std;const int MAXN = 100001*20, S = 10;struct SAM {    int chl[MAXN][S], fa[MAXN], maxl[MAXN];    int top, root, last;    void init()    {        top = root = last = 1;        memset(chl, 0, sizeof chl);        memset(fa, 0, sizeof fa);        memset(maxl, 0, sizeof maxl);    }    void push(int stat, int x)    {        int p = stat, np = ++top; maxl[np] = maxl[p] + 1;        while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];        if (!p) fa[np] = root;        else {            int q = chl[p][x];            if (maxl[q] == maxl[p] + 1) fa[np] = q;            else {                int nq = ++top; maxl[nq] = maxl[p] + 1;                memcpy(chl[nq], chl[q], sizeof chl[q]);                fa[nq] = fa[q], fa[q] = fa[np] = nq;                while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];            }        }        last = np;    }}sam;queue<int> que;int stat[102400], rd[102400], col[102400];struct node {    int to, next;} edge[202400];int head[102400], top = 0;void push(int i, int j){ rd[i]++, ++top, edge[top] = (node) {j, head[i]}, head[i] = top; }void bfs(int nd){    //printf("BFS : %d\n", nd);    memset(stat, 0, sizeof stat);    stat[nd] = 1;    que.push(nd);    while (!que.empty()) {        int tp = que.front(); que.pop();        sam.push(stat[tp], col[tp]);        //printf("%d -- %d--+%d-->%d\n", tp, stat[tp], col[tp], sam.last);        for (int i = head[tp]; i; i = edge[i].next)            if (!stat[edge[i].to])                stat[edge[i].to] = sam.last, que.push(edge[i].to);    }}int n, c;void solve(){    sam.init();    scanf("%d%d", &n, &c);    for (int i = 1; i <= n; i++)        scanf("%d", &col[i]);    for (int i = 1; i < n; i++) {        int u, v; scanf("%d%d", &u, &v);        push(u, v); push(v, u);    }    for (int i = 1; i <= n; i++)        if (rd[i] == 1)            bfs(i);    long long ans = 0;    for (int i = 2; i <= sam.top; i++)        ans += sam.maxl[i] - sam.maxl[sam.fa[i]];    printf("%lld", ans);}int main(){    //freopen("zjoi15_substring.in", "r", stdin);    //freopen("zjoi15_substring.out", "w", stdout);    solve();    return 0;}

POI2000 公共串

给你n≤5个长度不超过2000的字符串，求最长公共子串。

SAM解法：只要在匹配的时候记录每个节点对于第i个串匹配的最长距离，然后xjb取max和min就好了。

APIO2014 回文串

给定一个字符串S，求一个回文子串T，最大化 |T|∗times(T)，times(T) 为出现次数。

首先我们用manacher算法求出本质不同的回文串。由于manacher的复杂度为O(n)，回文串个数为O(n)。我们一个个询问其出现次数即可。现在问题转化为了对于一个子串，要在 O(lgn) 的时间内求出其出现次数。

首先我们预处理出pos[r]:=S[1..r]在后缀自动机中的位置，那么 S[l,r] 就是parent树上最后一个maxl大于l-r+1的节点。用倍增处理即可。

#include <bits/stdc++.h>using namespace std;const int MAXN = 300005*2, S = 26;int pos[MAXN]; // S[1..r]对应状态int fa[MAXN][21];char str[MAXN];int right_siz[MAXN];int stk[MAXN], top = 0, rd[MAXN];struct SAM {    int chl[MAXN][S], fa[MAXN], maxl[MAXN];    int top, root, last;    void clear()    {        top = root = last = 1;        memset(chl, 0, sizeof chl), memset(fa, 0, sizeof fa), memset(maxl, 0, sizeof maxl);    }    SAM() { clear(); }    void push(int x)    {        int p = last, np = ++top; maxl[np] = maxl[p] + 1, right_siz[np]++;        while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];        if (!p) fa[np] = root;        else {            int q = chl[p][x];             if (maxl[q] == maxl[p] + 1) fa[np] = q;            else {                int nq = ++top; maxl[nq] = maxl[p]+1;                memcpy(chl[nq], chl[q], sizeof chl[q]);                fa[nq] = fa[q], fa[q] = fa[np] = nq;                while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];            }        }        last = np;    }} sam;void top_sort(){    for (int i = 1; i <= sam.top; i++) rd[sam.fa[i]]++;    for (int i = 1; i <= sam.top; i++) if (rd[i] == 0) stk[++top] = i;    while (top) {        int t = stk[top--];  rd[sam.fa[t]]--, right_siz[sam.fa[t]] += right_siz[t];        if (rd[sam.fa[t]] == 0) stk[++top] = sam.fa[t];    }}void init(){    scanf("%s", str+1);    for (char *p = str+1; *p != '\0'; ++p)        sam.push(*p-'a');    int len = strlen(str+1);    for (int i = 1, nd = sam.root; i <= len; i++) {        nd = sam.chl[nd][str[i]-'a'];        pos[i] = nd;    }    for (int i = 1; i <= sam.top; i++) fa[i][0] = sam.fa[i];    for (int j = 1; j <= 20; j++)        for (int i = 1; i <= sam.top; i++)            fa[i][j] = fa[fa[i][j-1]][j-1];    top_sort(); // Count right_siz} long long ans = 0;void query(int i, int j){    int nd = pos[j];    for (int k = 20; k >= 0; k--)        if (sam.maxl[fa[nd][k]] >= j-i+1)            nd = fa[nd][k];    ans = max(ans, (long long)(j-i+1)*right_siz[nd]);}int p[MAXN];void work(){    int len = strlen(str+1);    int id = 0, mx = 0; // manacher    str[0] = '$';     for (int i = 1; i <= len; i++) {        if (mx > i) p[i] = min(p[id-(i-id)], mx-i); else p[i] = 1, query(i, i);        while (str[i-p[i]] == str[i+p[i]]) query(i-p[i], i+p[i]), p[i]++;        if (i+p[i] > mx) id = i, mx = i+p[i];     }    id = mx = 0;    for (int i = 1; i <= len; i++) {        if (mx > i) p[i] = min(p[id-(i-id)], mx-i); else p[i] = 0;        while (str[i-p[i]] == str[i+p[i]+1]) query(i-p[i], i+p[i]+1), p[i]++;        if (i+p[i] > mx) id = i, mx = i+p[i];    }    cout << ans << endl;}int main(){    init();    work();    return 0;}

HAOI2016找相同字符

给定两个串S1，S2，统计他们的公共子串总数。两个子串不同，当且仅当长度不同或出现位置不同。

SA解法：将S1和S2用一个’#’隔开，求出height数组，由于公共子串是后缀的前缀，因此答案就是所有前一半的后缀和后一半的后缀的lcp的和。用单调栈扫两遍记录答案即可。最优复杂度O(n)。

SAM解法：这个做法比较鬼畜。先把第一个串建立后缀自动机，再把第二个串在上面跑。到达一个状态x时匹配长度为len对答案的贡献分为两部分：

1. 当前位置的收获 (len−min(x)+1)×|Right(x)|
2. 由于匹配了当前位置，自然匹配了父亲节点，就有 ∑i为x的祖先(max(i)−min(i)+1)×|Right(i)|

第一部分为O(1)，第二部分可以拓扑排序后 O(n) 预处理。总复杂度为O(n)。

#include <bits/stdc++.h>using namespace std;const int MAXN = 200005*2, S = 26;int right_siz[MAXN];struct SAM {    int chl[MAXN][S], fa[MAXN], maxl[MAXN];    int top, root, last;    void clear()     { top = root = last = 1; }    SAM()    { clear(); }    void push(int x)    {        int p = last, np = ++top; maxl[np] = maxl[p] + 1; right_siz[np]++;        while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];        if (!p) fa[np] = root;        else {            int q = chl[p][x];            if (maxl[q] == maxl[p] + 1) fa[np] = q;            else {                int nq = ++top; maxl[nq] = maxl[p] + 1;                memcpy(chl[nq], chl[q], sizeof chl[q]);                fa[nq] = fa[q], fa[q] = fa[np] = nq;                while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];            }        }        last = np;    }} sam;char s1[MAXN], s2[MAXN];int stk[MAXN], top = 0;int rd[MAXN];int topo[MAXN], tp_top = 0;int canc[MAXN];int vis[MAXN];void dfs(int nd, string str) {    printf("Id = %d, pre = %d, dis = %d, right = %d, cacc = %d\n", nd, sam.fa[nd], sam.maxl[nd], right_siz[nd], canc[nd]);    vis[nd] = 1;    for (int i = 0; i < S; i++)        if (sam.chl[nd][i])            printf("-+%c-> %d\n", i+'a', sam.chl[nd][i]);    for (int i = 0; i < S; i++)        if (sam.chl[nd][i] && !vis[sam.chl[nd][i]])            dfs(sam.chl[nd][i], str+char(i+'a'));}void top_sort(){    for (int i = 1; i <= sam.top; i++) rd[sam.fa[i]]++;    for (int i = 1; i <= sam.top; i++) if (rd[i] == 0) stk[++top] = i;    while (top) {        int t = stk[top--]; topo[++tp_top] = t, rd[sam.fa[t]]--;        if (rd[sam.fa[t]] == 0) stk[++top] = sam.fa[t];    }    for (int i = 1; i <= tp_top; i++) right_siz[sam.fa[topo[i]]] += right_siz[topo[i]];    for (int i = tp_top; i >= 1; i--)         if (topo[i] != sam.root && sam.fa[topo[i]] != sam.root)             canc[topo[i]] = canc[sam.fa[topo[i]]] + (sam.maxl[sam.fa[topo[i]]]-sam.maxl[sam.fa[sam.fa[topo[i]]]])*right_siz[sam.fa[topo[i]]];}void work(){    scanf("%s%s", s1, s2);    for (char *p = s1; *p != '\0'; p++) sam.push(*p-'a');    top_sort();    int nd = sam.root, len = 0;    long long ans = 0;    //int l = strlen(s2); s2[l] = '$', s2[l+1] = '\0';    for (char *p = s2; *p != '\0'; p++) {        if (sam.chl[nd][*p-'a']) nd = sam.chl[nd][*p-'a'], len++;        else {            while (nd && !sam.chl[nd][*p-'a']) nd = sam.fa[nd];            if (!nd) nd = sam.root, len = 0;            else len = sam.maxl[nd]+1, nd = sam.chl[nd][*p-'a'];        }        ans += canc[nd] + (len-sam.maxl[sam.fa[nd]])*right_siz[nd];        //cout << ans << endl;    }    cout << ans << endl;}int main(){    work();    return 0;}

2555: SubString

LCT维护Right数组大小…

神题。

#include <bits/stdc++.h>using namespace std;const int MAXN = 1600002, S = 26;struct LCT {    int chl[MAXN][2], fa[MAXN], siz[MAXN], flag[MAXN], rev[MAXN], add[MAXN];    int stk[MAXN];    int root, top;    void clear()    { root = top = 0; }    LCT()    { clear(); }    bool isrt(int nd)    { return chl[fa[nd]][0] != nd && chl[fa[nd]][1] != nd; }    void pdw(int nd)    {        int &lc = chl[nd][0], &rc = chl[nd][1];        if (lc) rev[lc] ^= rev[nd], add[lc] += add[nd];        if (rc) rev[rc] ^= rev[nd], add[rc] += add[nd];        if (rev[nd]) rev[nd] = 0, swap(lc, rc);        if (add[nd]) siz[nd] += add[nd], add[nd] = 0;    }    void zig(int nd)    {        int p = fa[nd], g = fa[p];        int tp = chl[p][0] != nd, tg = chl[g][0] != p, son = chl[nd][tp^1];        if (!isrt(p)) chl[g][tg] = nd;        chl[nd][tp^1] = p, chl[p][tp] = son;        fa[nd] = g, fa[p] = nd, fa[son] = p;    }    void splay(int nd)    {        int top = 0; stk[++top] = nd;        for (int x = nd; !isrt(x); x = fa[x])            stk[++top] = fa[x];        while (top) pdw(stk[top--]);        while (!isrt(nd)) {            int p = fa[nd], g = fa[p];            int tp = chl[p][0] != nd, tg = chl[g][0] != p;            if (isrt(p)) { zig(nd); break; }            else if (tp == tg) zig(p), zig(nd);            else zig(nd), zig(nd);        }    }    void dfs(int nd, int tab)    {        if (!nd) return;        for (int i = 1; i <= tab; i++) putchar(' ');        printf("nd = %d, flag = %d, siz = %d, lc = %d, rc = %d, fa = %d, rev = %d\n", nd, flag[nd], siz[nd], chl[nd][0], chl[nd][1], fa[nd], rev[nd]);        dfs(chl[nd][0], tab+2);        dfs(chl[nd][1], tab+2);    }    void access(int x)    {        for (int y = 0; x; x = fa[y = x])            splay(x), chl[x][1] = y;    }    void mkt(int x)    { access(x), splay(x), rev[x] ^= 1; }    void link(int x, int y)    { mkt(x); splay(x); fa[x] = y; }    void cut(int x, int y)    { mkt(x), access(y), splay(y), fa[x] = chl[y][0] = 0;}    void lct_link(int x, int y) // x->y    {        //printf("LINK : %d-->%d\n", x, y);        link(x, y), mkt(1);        //puts("---");        access(y), splay(y), siz[y] += siz[x];        //puts("---");        if (chl[y][0]) add[chl[y][0]] += siz[x];    }    void lct_cut(int x, int y) // cut x->y    {        cut(x, y), mkt(1);        access(y), splay(y), siz[y] -= siz[x];        if (chl[y][0]) add[chl[y][0]] -= siz[x];    }    void set_flag(int x)    { mkt(x), splay(x), siz[x] = 1; }    int find_fa(int x)    {        access(x);        while (!isrt(x)) x = fa[x];        return x;    }    int query(int nd)    {        mkt(nd), splay(nd);        return siz[nd];    }} lct;struct SAM {    int chl[MAXN*2][S], fa[MAXN*2], maxl[MAXN*2];    int top, last, root;    void clear()    { top = last = root = 1; }    SAM()    { clear(); }    void push(int x)    {        //cout << "PUSH : " << (char)(x+'a') << endl;        int p = last, np = ++top; maxl[np] = maxl[p] + 1; lct.set_flag(np);        //puts("j");        while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];        //puts("jj");        if (!p) fa[np] = root, lct.lct_link(np, root);        else {            int q = chl[p][x];            if (maxl[q] == maxl[p] + 1) fa[np] = q, lct.lct_link(np, q);            else {                int nq = ++top; maxl[nq] = maxl[p] + 1;                memcpy(chl[nq], chl[q], sizeof chl[q]);                lct.lct_link(nq, fa[q]), fa[nq] = fa[q];                lct.lct_cut(q, fa[q]), lct.lct_link(q, nq), fa[q] = nq;                lct.lct_link(np, nq), fa[np] = nq;                while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];            }        }        //puts("jjj");        last = np;    }} sam;char str[MAXN*2];int q, mask = 0;void decode(int mask){    int len = strlen(str);    for (int j = 0; j < len; j++) {        mask = (mask*131+j)%len;        swap(str[j], str[mask]);    }}void get_str(char str[]){    scanf("%s", str);    decode(mask);}char opt[10];int main(){    //freopen("substring.in", "r", stdin);    //freopen("substring.out","w",stdout);    scanf("%d", &q);    scanf("%s", str);    //puts("hah");    for (char *p = str; *p != '\0'; p++)        sam.push(*p-'A');    //puts("hah");    for (int i = 1; i <= q; i++) {        scanf("%s", opt);        //cout << opt << endl;        if (opt[0] == 'A') {            get_str(str);            for (char *p = str; *p != '\0'; p++)                sam.push(*p-'A');        } else {            get_str(str);            int nd = sam.root, flag = 0;            for (char *p = str; *p != '\0'; p++) {                if (!sam.chl[nd][*p-'A']) {flag = 1; break; }                else nd = sam.chl[nd][*p-'A'];            }            if (flag) puts("0");            else {                int ans = lct.query(nd);                printf("%d\n", ans);                mask ^= ans;            }        }    }    return 0;}

poj2774: Long Long Message

裸题，SAM直接碾。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 200005;int chl[MAXN][26], fa[MAXN], maxl[MAXN];int top = 1, root = 1, last = 1;void push(int x){    int p = last, np = ++top; maxl[np] = maxl[p]+1;    while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];    if (!p) fa[np] = root;    else {        int q = chl[p][x];        if (maxl[q] == maxl[p]+1) fa[np] = q;        else {            int nq = ++top; maxl[nq] = maxl[p]+1;            memcpy(chl[nq], chl[q], sizeof chl[q]);            fa[nq] = fa[q], fa[q] = fa[np] = nq;            while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];        }    }    last = np;}char str[MAXN];int main(){    scanf("%s", str);    for (char *p = str; *p != '\0'; ++p)        push(*p-'a');    scanf("%s", str);    int nd = root, len = 0, ans = 0;    for (char *p = str; *p != '\0'; ++p) {        int x = *p-'a';        if (chl[nd][x]) nd = chl[nd][x], len++;        else {            while (nd && !chl[nd][x]) nd = fa[nd];            if (!nd) nd = root, len = 0;            else len = maxl[nd]+1, nd = chl[nd][x];        }        ans = max(ans, len);    }    cout << ans << endl;    return 0;}

[SPOJ705]不同的子串

模板复习计划，裸题。

#include <bits/stdc++.h>using namespace std;const int MAXN = 50005*2;struct SAM {    int chl[MAXN][26], fa[MAXN], maxl[MAXN];    int top, root, last;    void clear()    { top = root = last = 1, memset(chl, 0, sizeof chl),    memset(fa, 0, sizeof fa), memset(maxl, 0, sizeof maxl); }    SAM() { clear(); }    void push(int stat, int x)    {        int p = last, np = ++top; maxl[np] = maxl[p]+1;        while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];        if (!p) fa[np] = root;        else {            int q = chl[p][x];            if (maxl[q] == maxl[p]+1) fa[np] = q;            else {                int nq = ++top; maxl[nq] = maxl[p]+1;                memcpy(chl[nq], chl[q], sizeof chl[q]);                fa[nq] = fa[q], fa[q] = nq, fa[np] = nq;                while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];            }        }        last = np;    }} sam;char str[MAXN];int main(){    freopen("subst1.in", "r", stdin);    freopen("subst1.out", "w", stdout);    scanf("%s", str+1);    for (char *p = str+1; *p != '\0'; ++p)        sam.push(sam.last, *p-'A');    long long ans = 0;    for (int i = 1; i <= sam.top; i++)        ans += sam.maxl[i]-sam.maxl[sam.fa[i]];    cout << ans << endl;    return 0;}

bzoj2806: [Ctsc2012]Cheat

比较神的题…
后缀自动机上dp，由于dp决策有区间性质，可以用线段树或者单调队列维护。线段树版本(O(nlg2n))会TLE，明天写单调队列的。

线段树(TLE):

#include <bits/stdc++.h>using namespace std;const int MAXN = 1100001*2;int tree[MAXN][2], fin[MAXN], col[MAXN], trie_root = 0, trie_top = 0;void push_str(int &nd, const char *str){    if (!nd) nd = ++trie_top;    if (*str == '\0') fin[nd] = 1;    else push_str(tree[nd][*str-'0'], str+1), col[tree[nd][*str-'0']] = *str-'0';}int chl[MAXN][2], fa[MAXN], maxl[MAXN], root = 1, top = 1;void push(int p, int x, int &last){    int np = ++top; maxl[np] = maxl[p]+1;    while (p && !chl[p][x]) chl[p][x] = np, p = fa[p];    if (!p) fa[np] = root;    else {        int q = chl[p][x];        if (maxl[q] == maxl[p]+1) fa[np] = q;        else {            int nq = ++top; maxl[nq] = maxl[p]+1;            chl[nq][0] = chl[q][0], chl[nq][1] = chl[q][1];            fa[nq] = fa[q], fa[q] = fa[np] = nq;            while (p && chl[p][x] == q) chl[p][x] = nq, p = fa[p];        }    }    last = np;}queue<int> que;int stat[MAXN];void build_tree(){    que.push(trie_root), stat[trie_root] = 1;    while (!que.empty()) {        int tp = que.front(); que.pop();        int last;        push(stat[tp], col[tp], last);        for (int i = 0; i <= 1; i++)            if (tree[tp][i])                stat[tree[tp][i]] = last, que.push(tree[tp][i]);    }}bool match(char str[], int l, int r){    if (l <= 0 || l > r) return 0;    int nd = root, ans = 0, len = 0;    for (int i = l; i <= r; i++) {        int x = str[i]-'0';        if (chl[nd][x]) nd = chl[nd][x], len++;        else {            while (nd && !chl[nd][x]) nd = fa[nd];            if (!nd) nd = root, len = 0;            else len = maxl[nd]+1, nd = chl[nd][x];        }        ans = max(ans, len);    }     return ans == r-l+1;}char str[MAXN];int dp[MAXN], max_back[MAXN];int n, m;int zkw[(1<<21)+1], N = 1<<20;void modify(int nd, int val){    nd += N-1;    zkw[nd] = val;    for (int i = nd>>1; i; i >>= 1)        zkw[i] = max(zkw[i*2], zkw[i*2+1]);}int ask_max(int l, int r){    if (l > r) return 0;    int ans = -233333333;    for (l += N-1, r += N-1; l < r; l>>=1, r>>=1) {        if (l&1) ans = max(ans, zkw[l++]);        if (!(r&1)) ans = max(ans, zkw[r--]);    }    if (l == r) ans = max(ans, zkw[l]);    return ans;}bool do_dp(int L){    int l = strlen(str+1);    memset(dp, 0, sizeof dp);    memset(zkw, -127/3, sizeof zkw);    int max_val = 0;    modify(0, 0);    for (int i = 1; i <= l; i++) {        dp[i] = max_val;        if (i-L >= 0 && max_back[i] <= i-L)            dp[i] = max(dp[i], i+ask_max(max_back[i], i-L));        max_val = max(max_val, dp[i]);        modify(i, dp[i]-i);    }    return dp[l]*10 >= l*9;}int main(){    scanf("%d%d", &n, &m);    for (int i = 1; i <= m; i++) {        scanf("%s", str);        push_str(trie_root, str);    }    build_tree();    for (int i = 1; i <= n; i++) {        scanf("%s", str+1);        memset(max_back, 0, sizeof max_back);        for (int i = 1, l = strlen(str+1); i <= l; i++) {            int lf = 1, rt = i, mid;            while (lf <= rt) {                mid = (lf+rt)>>1;                if (match(str, i-mid+1, i)) lf = mid+1;                else rt = mid-1;            }            max_back[i] = i-(lf-1);        }         int l = 1, r = strlen(str+1), mid;        while (l <= r) {            mid = (l+r)>>1;            if (do_dp(mid)) l = mid+1;            else r = mid-1;        }        printf("%d\n", l-1);    }    return 0;}