HDU 6138 Fleet of the Eternal Throne(后缀自动机)

题意

题目链接

Sol

真是狗血，被疯狂卡常的原因竟是

我们考虑暴力枚举每个串的前缀，看他能在(x, y)的后缀自动机中走多少步，对两者取个min即可

复杂度(O(T 10^5 M))(好假啊)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
int N, M;
string s[MAXN];
struct SAM {
	int ch[MAXN][26], fa[MAXN], len[MAXN], tot, las, root;
	void init() {
		for(int i = 0; i <= tot; i++) 
			fa[i] = 0, len[i] = 0, memset(ch[i], 0, sizeof(ch[i]));
		tot = root = 1; las = 1;
	}
	void insert(int x) {
		int now = ++tot, pre = las; las = now; len[now] = len[pre] + 1;
		for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
		if(!pre) {fa[now] = root; return ;}
		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; fa[q] = fa[now] = nq;
			memcpy(ch[nq], ch[q], sizeof(ch[q]));
			for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
		}
	}
	void Build(string str) {
		init();	
		for(auto &x: str) 
			insert(x - 'a');
	}
	int find(string str) {
		int cur = 0, now = root;
		for(auto &x : str) {
			int v = x - 'a';
			if(ch[now][v]) cur++, now = ch[now][v];
			else return cur;
		}
		return cur;
	}
}S[2];


void solve() {
	cin >> N;
	for(int i = 1; i <= N; i++) cin >> s[i];
	cin >> M;
	while(M--) {
		int x, y;
		cin >> x >> y;
		S[0].Build(s[x]);
		S[1].Build(s[y]);
		int ans = 0;
		for(int i = 1; i <= N; i++) 
			ans = max(ans, min(S[0].find(s[i]), S[1].find(s[i])));
		cout << ans << '
';
	}	
}
int main() {
//	freopen("a.in", "r", stdin);
	ios::sync_with_stdio(0);
	int T; cin >> T;
    for(; T--; solve());
    return 0;
}