后缀自动姬

是什么？

博主不想写~~反正没人看~~

直接去\(hihocoder\)

模板

# include 
    
     # define IL inline# define RG register# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;template 
     
      IL void Input(RG Int &x){    RG int z = 1; RG char c = getchar(); x = 0;    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    x *= z;}const int maxn(2e6 + 5);int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1;char s[maxn];IL void Extend(RG int c){    RG int np = ++tot, p = last; last = np;    len[np] = len[p] + 1;    while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];    if(!p) fa[np] = 1;    else{        RG int q = trans[c][p];        if(len[p] + 1 == len[q]) fa[np] = q;        else{            RG int nq = ++tot;            len[nq] = len[p] + 1, fa[nq] = fa[q];            for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];            fa[q] = fa[np] = nq;            while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];        }    }}int main(RG int argc, RG char* argv[]){    scanf(" %s", s), n = strlen(s);    for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');    return 0;}

用法

一个串不同子串的个数

就是\(endpos(right)\)集合的大小和

就是\(\sum_{i}len[i]-len[fa[i]]\)

两个串的最长公共子串长度

一个串建立\(sam\)

另一个串匹配就好啦

for(RG int i = 0, nw = 1, cnt = 0; i < n; ++i){    RG int c = s[i] - 'a';    if(trans[c][nw]) ++cnt, nw = trans[c][nw];    else{        while(nw && !trans[c][nw]) nw = fa[nw];        if(!nw) nw = 1, cnt = 0;        else cnt = len[nw] + 1, nw = trans[c][nw];    }    ans = max(ans, cnt);}

拓扑序的快速求法

对\(len\)来个桶排序

for(RG int i = 1; i <= tot; ++i) ++t[len[i]];for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;

多个串的最长公共子串

一个串建立\(sam\)

其它的串在上面匹配

可以类似\(DP\)的在拓扑序上转移

for(RG int i = 1; i <= tot; ++i) ++t[g[i] = len[i]];for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;while(scanf(" %s", s) != EOF){    n = strlen(s);    for(RG int i = 1; i <= tot; ++i) f[i] = 0;    for(RG int i = 0, nw = 1, cnt = 0; i < n; ++i){        RG int c = s[i] - 'a';        if(trans[c][nw]) ++cnt, nw = trans[c][nw];        else{            while(nw && !trans[c][nw]) nw = fa[nw];            if(!nw) nw = 1, cnt = 0;            else cnt = len[nw] + 1, nw = trans[c][nw];        }        f[nw] = max(f[nw], cnt);    }    for(RG int i = tot; i; --i) f[fa[id[i]]] = max(f[fa[id[i]]], f[id[i]]);    for(RG int i = 1; i <= tot; ++i) g[i] = min(g[i], f[i]);}for(RG int i = 1; i <= tot; ++i) ans = max(ans, g[i]);