洛谷P3346 [ZJOI2015]诸神眷顾的幻想乡(广义后缀自动机)

题意

题目链接

Sol

广义SAM的板子题。

首先叶子节点不超过20，那么可以直接对每个叶子节点为根的子树插入到广义SAM中。

因为所有合法的答案一定是某个叶子节点为根的树上的一条链，因此这样可以统计出所有合法的答案

然后就是经典的本质不同子串问题了，\(ans = \sum len[i] - len[fa[i]]\)

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 2e6 + 10;
inline int read() {
	char c = getchar(); int x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
int N, C, a[MAXN], deg[MAXN];
namespace SAM {
	int fa[MAXN], len[MAXN], ch[MAXN][11], tot = 1, root = 1, las = 1;
	int insert(int x, int pre) {
		int now = ++tot; len[now] = len[pre] + 1;
		for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
		if(!pre) {fa[now] = root; return now;}
		int q = ch[pre][x];
		if(len[q] == len[pre] + 1) fa[now] = q;
		else {
			int nq = ++tot; fa[nq] = fa[q]; len[nq] = len[pre] + 1;
			memcpy(ch[nq], ch[q], sizeof(ch[q]));
			for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
			fa[q] = fa[now] = nq;
		}
		return now;
	}
	LL calc() {
		LL ans = 0;
		for(int i = 1; i <= tot; i++) ans += len[i] - len[fa[i]];
		return ans;
	}
}
vector<int> v[MAXN], node;
void dfs(int x, int _fa, int p) {
	p = SAM::insert(a[x], p);
	for(auto &to : v[x]) {
		if(to == _fa) continue;
		dfs(to, x, p);
	}
}
int main() {
	N = read(); C = read();
	for(int i = 1; i <= N; i++) a[i] = read();
	for(int i = 1; i <= N - 1; i++) {
		int x = read(), y = read();
		v[x].push_back(y); deg[x]++;
		v[y].push_back(x); deg[y]++;
	}
	for(int i = 1; i <= N; i++)
		if(deg[i] == 1) SAM::las = 1, dfs(i, 0, 1);
	cout << SAM::calc();
	return 0;
}