SPOJ: LCS

题面

输入(2)个长度不大于(250000)的字符串，输出这(2)个字符串的最长公共子串。如果没有公共子串则输出(0)

Sol

一个串建立(sam)
另一个串在上面匹配

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

template <class Int>
IL void Input(RG Int &x){
	RG int z = 1; RG char c = getchar(); x = 0;
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	x *= z;
}

const int maxn(5e5 + 5);

int n, trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1, ans;
char s[maxn];

IL void Extend(RG int c){
	RG int np = ++tot, p = last; last = np;
	len[np] = len[p] + 1;
	while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
	if(!p) fa[np] = 1;
	else{
		RG int q = trans[c][p];
		if(len[q] == len[p] + 1) fa[np] = q;
		else{
			RG int nq = ++tot;
			len[nq] = len[p] + 1, fa[nq] = fa[q];
			for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
			fa[q] = fa[np] = nq;
			while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
		}
	}
}

int main(RG int argc, RG char* argv[]){
	scanf(" %s", s), n = strlen(s);
	for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
	scanf(" %s", s), n = strlen(s);
	for(RG int i = 0, nw = 1, cnt = 0; i < n; ++i){
		RG int c = s[i] - 'a';
		if(trans[c][nw]) ++cnt, nw = trans[c][nw];
		else{
			while(nw && !trans[c][nw]) nw = fa[nw], cnt = len[nw];
			if(!nw) nw = 1, cnt = 0;
			else cnt++, nw = trans[c][nw];
		}
		ans = max(ans, cnt);
	}
	printf("%d
", ans);
	return 0;
}