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\n", ans);

	}

	return 0;

}