后缀自动机学习

1. hihocoder #1441 : 后缀自动机一·基本概念

按照后缀自动机概念模拟即可, 复杂度$O(n^3logn)$.

#include <iostream>
#include <map>
#include <string>
#include <set>
using namespace std;

const int N = 111;
int n, cnt;
string s;
map<string,set<int> > suf_to_endpos;
map<set<int>,int> endpos_id;
map<string,int> suf_id;
set<string> st[N];
string longest[N], shortest[N];
set<int> endpos[N];

int getID(set<int> &s) {
	if (endpos_id.count(s)) return endpos_id[s];
	endpos[++cnt] = s;
	return endpos_id[s] = cnt;
} 

int main() {
	cin>>s>>n;
	for (int i=0; i<s.size(); ++i) {
		for (int j=1; i+j<s.size(); ++j) {
			suf_to_endpos[s.substr(i,j)].insert(i+j);
		}
	}
	for (auto &t:suf_to_endpos) { 
		int id = getID(t.second);
		string suf = t.first;
		st[id].insert(suf);
		suf_id[suf] = id;
		if (longest[id].empty()) longest[id] = shortest[id] = suf;
		else {
			if (suf.size()>longest[id].size()) longest[id] = suf;
			if (suf.size()<shortest[id].size()) shortest[id] = suf;
		}
	}
	while (n--) {
		cin>>s;
		int id = suf_id[s];
		cout<<shortest[id]<<' '<<longest[id];
		for (auto &&t:endpos[id]) cout<<' '<<t;
		cout<<endl;
	}
}

2. hihocoder #1445 : 后缀自动机二·重复旋律5