[SPOJ LCS]Longest Common Substring && 后缀自动机-优快云博客

本文链接：https://blog.youkuaiyun.com/shiyukun1998/article/details/46382531

本文详细介绍了SAM（Suffix Array with Modification）字符串匹配算法的实现原理及应用。通过C++代码展示了如何构建SAM树并利用其解决字符串匹配问题，包括关键步骤如节点扩展、回溯等。

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又来码一遍模板题~.~

然后还码错啦 fa[nq] = fa[q] 打成了fa[nq] = p

#include<cstdio>  
#include<algorithm>  
#include<cstring>  
#include<iostream>  
#include<queue>  
#define SF scanf  
#define PF printf  
#define idx(c) (c-'a')  
using namespace std;  
typedef long long LL;  
const int MAXN = 250000;
int ncnt, last;    
struct SAM {
	int ch[MAXN*2+10][26], fa[MAXN*2+10], step[MAXN*2+10];
	int last, ncnt;
	SAM () { last = ++ncnt; }
	void extend(int c) {
		int p = last, np = last = ++ncnt;
		step[np] = step[p]+1;
		while(!ch[p][c] && p) ch[p][c] = np, p = fa[p];
		if(!p) fa[np] = 1;
		else {
			int q = ch[p][c], nq;
			if(step[q] == step[p]+1) fa[np] = q;
			else {
				nq = ++ncnt;
				step[nq] = step[p]+1;
				memcpy(ch[nq], ch[q], sizeof(ch[q]));
				fa[nq] = fa[q]; fa[q] = nq; fa[np] = nq;
				while(ch[p][c] == q && p) ch[p][c] = nq, p = fa[p];
			}
		}
	}
	void build(char *s) {
		int len = strlen(s);
		for(int i = 0; i < len; i++) extend(idx(s[i]));
	}
	void solve(char *s) { 
		int len = strlen(s);
		int cur = 0, ans = 0, u = 1;
		for(int i = 0; i < len; i++) {
			int c = idx(s[i]);
			if(ch[u][c]) {
				cur++;
				u = ch[u][c];
			}
			else {
				while(!ch[u][c] && u) u = fa[u];
				if(!u)
					u = 1, cur = 0;
				else 
					cur = step[u]+1, u = ch[u][c];
			}
			ans = max(ans, cur);
		}
		PF("%d\n", ans);
	}
} ac;
char s1[MAXN+10], s2[MAXN+10];
int main() {
	SF("%s%s", s1, s2);
	ac.build(s1);
	ac.solve(s2);
	return 0;
}