[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=3926

[算法]

建立廣義後綴自動機

對於每個葉子節點 ， 以它為根 ， 依次將路徑上的子串加入自動機

最後統計本質不同的子串個數即可

時間復雜度 : O(N)

[代碼]

#include<bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
typedef long long ll;
typedef long double
 
 ld;
typedef unsigned long long ull;

int n , m;
int val[N];
vector< int > a[N];

struct Suffix_Automaton
{
        int size;
        int father[N << 1] , child[N << 1][15] , depth[N << 1];
        Suffix_Automaton()
        {
                size = 1;
        }
        inline 
 
int new_node(int dep)
        {
                depth[++size] = dep;
                memset(child[size] , 0 , sizeof(child[size]));
                father[size] = 0;
                return size;
        }
        inline int extend(int last , int ch) 
        {
                int np = child[last][ch];
                
 
if (np) 
                {    
                    if (depth[np] == depth[last] + 1) return np;
                    else {
                        int nq = new_node(depth[last] + 1);
                        father[nq] = father[np];
                        father[np] = nq;
                        memcpy(child[nq], child[np], sizeof(child[np]));
                        for (int p = last; child[p][ch] == np; p = father[p])
                            child[p][ch] = nq;
                        return nq;
                    }
                } else 
                {
                    np = new_node(depth[last] + 1);
                    int p = last;
                    for (; child[p][ch] == 0; p = father[p])
                        child[p][ch] = np;
                    if (child[p][ch] == np) 
                    {
                            father[np] = 1;
                            return np;
                    }
                    int q = child[p][ch];
                    if (depth[p] + 1 == depth[q]) 
                    {
                            father[np] = q;
                            return np;
                    } else 
                    {
                            int nq = new_node(depth[p] + 1);
                            father[nq] = father[q];
                            father[q] = father[np] = nq;
                            memcpy(child[nq], child[q], sizeof(child[q]));
                            for (; child[p][ch] == q; p = father[p])
                                    child[p][ch] = nq;
                            return np;
                    }
                }
        }
        inline ll calc()
        {
                ll ans = 0;
                for (int i = 2; i <= size; i++) ans += depth[i] - depth[father[i]];
                return ans;
        }
} SAM;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline void dfs(int u , int par , int pre)
{
        pre = SAM.extend(pre , val[u]);
        for (unsigned i = 0; i < a[u].size(); i++) 
        {
                int v = a[u][i];
                if (v != par) dfs(v , u , pre);    
        }        
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= n; i++) read(val[i]);
        for (int i = 1; i < n; i++)
        {
                int x , y;
                read(x); read(y);
                a[x].push_back(y);
                a[y].push_back(x);
        }
        for (int i = 1; i <= n; i++)
                if (a[i].size() == 1) dfs(i , 0 , 1);
        printf("%lld\n" , SAM.calc());
                
        return 0;
    
}

[ZJOI 2015] 諸神眷顧的幻想鄉