【BZOJ3926】【ZJOI2015】诸神眷顾的幻想乡 广义后缀自动机

题目：

题目在这里

思路&做法：

参考的题解

既然只有\(20\)个叶子节点， 那么可以从每个叶子节点往上建一颗\(trie\)树, 然后合并成一棵大的\(trie\)树， 然后构建广义后缀自动机。
\((\) 实现起来就是dfs时把节点上的颜色加进广义后缀自动机\()\)

再然后就统计一下就可以了\((ans += len[x] - len[fail[x])\)

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

const int N = 2000010;

long long Ans;

struct Suffix_Automaton
{   int nxt[N][10], fail[N], sz;
	int len[N];
	int root;
	
	Suffix_Automaton() { }
	
	inline int newnode(int l)
	{   memset(nxt[sz], 0, sizeof(nxt[sz]));
		fail[sz] = 0;
		len[sz] = l;
		return sz++;
	}
	
	inline void init()
	{   sz = 1;
		root = newnode(0);
	}
	
	int add(int c, int last)
	{   if(nxt[last][c])
		{	int p = last, q = nxt[p][c];
			if(len[q] == len[p]+1)
				return q;
			else
			{   int u = newnode(len[p] + 1);
				for(int i=0; i<10; i++) nxt[u][i] = nxt[q][i];
				fail[u] = fail[q];
				fail[q] = u;
				while(nxt[p][c] == q)
				{	nxt[p][c] = u;
					p = fail[p];
				}
				return u;
			}
		}
		else
		{	int now = newnode(len[last] + 1);
			int p = last;
			while(p && !nxt[p][c])
			{	nxt[p][c] = now;
				p = fail[p];
			}
			if(!p) fail[now] = root;
			else
			{   int q = nxt[p][c];
				if(len[q] == len[p]+1)
					fail[now] = q;
				else
				{   int u = newnode(len[p] + 1);
					for(int i=0; i<10; i++) nxt[u][i] = nxt[q][i];
					fail[u] = fail[q];
					fail[q] = fail[now] = u;
					while(nxt[p][c] == q)
					{   nxt[p][c] = u;
						p = fail[p];
					}
				}
			}
			Ans += (long long) (len[now] - len[fail[now]]);
			return now;
		}
	}
} tzw;

int color[N];

int r[N];

struct edge
{   int from, to;
	edge() { }
	edge(int _1, int _2):from(_1), to(_2) { }
} edges[2*N];
int head[N], nxt[2*N], tot;

inline void init()
{   memset(head, -1, sizeof(head));
	tot = 0;
}

inline void Add_Edge(int x, int y)
{   edges[tot] = edge(x, y);
	nxt[tot] = head[x];
	head[x] = tot++;
	edges[tot] = edge(y, x);
	nxt[tot] = head[y];
	head[y] = tot++;
}

void dfs(int x, int fa, int last)
{   last = tzw.add(color[x], last);
	for(int i=head[x]; ~i; i=nxt[i])
	{   edge & e = edges[i];
		if(e.to != fa)
			dfs(e.to, x, last);
	}
}

int main()
{   int n, m;
	scanf("%d %d", &n, &m);
	for(int i=1; i<=n; i++)
		scanf("%d", &color[i]);
	tzw.init();
	init();
	for(int i=1; i<n; i++)
	{   int x, y;
		scanf("%d %d", &x, &y);
		Add_Edge(x, y);
		r[x]++, r[y]++;
	}
	for(int i=1; i<=n; i++)
		if(r[i] == 1) dfs(i, 0, tzw.root);
	printf("%lld\n", Ans);
	return 0;
}