SPOJ - 1811
输入两个字符串A和B，求最长公共子串，构造字符串A的后缀自动机，然后将字符串B从头开始在后缀自动机中匹配，能匹配到则继续往下，cnt++， 否则，一直往前跳，一直到找到能匹配该字符的后缀，或者跳回root.
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> #include <bitset> #define INF 0x3f3f3f3f #define eps 1e-6 #define PI 3.1415926 #define mod 1000000009 #define base 2333 using namespace std; typedef long long LL; const int maxn = 3e5 + 10; const int maxx = 1e3 + 10; inline void splay(int &v) { v=0;char c=0;int p=1; while(c<'0' || c >'9'){if(c=='-')p=-1;c=getchar();} while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();} v*=p; } struct Node { Node *pre, *nxt[26]; int step; void Clear() { step = 0; pre = NULL; memset(nxt, NULL, sizeof(nxt)); } } *root, *last; Node st[maxn<<1], *cur; char str1[maxn], str2[maxn]; void init() { cur = st; root = last = cur++; root->Clear(); } void extend(int w) { Node *np = cur++, *p = last; np->Clear(); np->step = p->step+1; while(p && !p->nxt[w]) p->nxt[w] = np, p = p->pre; if(p == NULL) np->pre = root; else { Node *q = p->nxt[w]; if(q->step == p->step+1) np->pre = q; else { Node *nq = cur++; memcpy(nq->nxt, q->nxt, sizeof(q->nxt)); nq->pre = q->pre; nq->step = p->step+1; np->pre = q->pre = nq; while(p && p->nxt[w] == q) p->nxt[w] = nq, p = p->pre; } } last = np; } void solve() { while(scanf("%s%s", str1, str2) != EOF) { init(); int len1 = strlen(str1), len2 = strlen(str2); for(int i = 0; i < len1; i++) extend(str1[i]-'a'); int cnt = 0, ans = 0; Node *now = root; for(int i = 0; i < len2; i++) { int x = str2[i]-'a'; if(now->nxt[x]) { now = now->nxt[x]; cnt++; } else { while(now && !now->nxt[x]) now = now->pre; if(now == NULL) { now = root; cnt = 0; } else { cnt = now->step+1; now = now->nxt[x]; } } if(cnt > ans) ans = cnt; } printf("%d\n", ans); } } int main() { //srand(time(NULL)); //freopen("kingdom.in","r",stdin); //freopen("kingdom.out","w",stdout); solve(); }