SPOJ LCS 后缀自动机

用后缀自动机求两个长串的最长公共子串，效果拔群。多样例的时候memset要去掉。

解题思路就是跟CLJ的一模一样啦。

#pragma warning(disable:4996)
#include<cstring>
#include<string>
#include<iostream>
#include<cmath>
#include<vector>
#include<algorithm>
#define maxn 250050
using namespace std;

struct State{
	State *suf, *go[26];
	int val;
	State() :suf(0), val(0){
		memset(go, 0, sizeof(go));
	}
}*root, *last;

State statePool[maxn * 2], *cur;

void init()
{
	cur = statePool;
	root = last = cur++;
}

void extend(int w)
{
	State *p = last, *np = cur++;
	np->val = p->val + 1;
	while (p&&!p->go[w]) p->go[w] = np, p = p->suf;
	if (!p) np->suf = root;
	else{
		State *q = p->go[w];
		if (p->val + 1 == q->val){
			np->suf = q;
		}
		else{
			State *nq = cur++;
			memcpy(nq->go, q->go, sizeof q->go);
			nq->val = p->val + 1;
			nq->suf = q->suf;
			q->suf = nq;
			np->suf = nq;
			while (p&&p->go[w] == q){
				p->go[w] = nq, p = p->suf;
			}
		}
	}
	last = np;
}

char stra[maxn], strb[maxn];

int main()
{
	while (~scanf("%s%s", stra, strb))
	{
		init();
		//memset(statePool, 0, sizeof(statePool));
		int lena = strlen(stra);
		for (int i = 0; i < lena; i++){
			extend(stra[i] - 'a');
		}
		int ans = 0;
		int lenb = strlen(strb);
		State *p = root;
		int len = 0;
		for (int i = 0; i < lenb; i++){
			if (p->go[strb[i] - 'a']){
				len++; ans = max(ans, len);
				p = p->go[strb[i] - 'a'];
			}
			else{
				while (p&&!p->go[strb[i] - 'a']){
					p = p->suf;
				}
				if (!p) {
					p = root;
					len = 0;
				}
				else{
					len = p->val + 1;
					ans = max(len, ans);
					p = p->go[strb[i] - 'a'];
				}
			}
		}
		printf("%d
", ans);
	}
	return 0;
}