題意: 求兩個串的最長公共子串

分析: 字尾陣列中最核心,應用中最常用的就是 $height$陣列, 根據字尾陣列,我們肯定要先把兩個串連起來, 最暴力的解法可以想到就是列舉第一串的所有位置 $i$

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <string>
#include <math.h>
#include <stack>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define rep(i, l, r) for(int i = l; i < r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define dbgln(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
#define dbg(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int>pii;

const int N = (int) 10000 * 2 + 11;
const int M = (int) 1e6 + 11;
const int MOD = (int) 1e9 + 7;
const int INF = (int) 0x3f3f3f3f;
const ll INFF = (ll) 0x3f3f3f3f3f3f3f3f;
/*-----------------------------------------------------------*/
int n, k;
int Rank[N + 1], sa[N + 1], height[N + 1], tmp[N + 1];
int MIN[N + 1][40];
bool cmp_sa(int i, int j){
	if(Rank[i] != Rank[j]) return Rank[i] < Rank[j];
	else {
		int ri = i + k <= n ? Rank[i + k] : -1;
		int rj = j + k <= n ? Rank[j + k] : -1;
		return ri < rj;
	}
}
void SA(string s){ // 排名為 1 - n ,其他都是0 - (n - 1)
	// sa , 0 - (n - 1)
	n = s.size();
	for(int i = 0; i <= n; i++){
		sa[i] = i; Rank[i] = (i < n) ? s[i] : -1;
	}
	for(k = 1; k <= n; k <<= 1){
		sort(sa, sa + n + 1, cmp_sa);
		tmp[sa[0]] = 0;
		for(int i = 1; i <= n; i++){
			tmp[sa[i]] = tmp[sa[i - 1]] + (cmp_sa(sa[i - 1], sa[i]) ? 1 : 0);
		}
		for(int i = 0; i <= n; i++){
			Rank[i] = tmp[i];
		}
	}

	// height and rank
	for(int i = 0; i <= n; i++) Rank[sa[i]] = i;
	int h = 0;
	height[0] = 0;
	for(int i = 0; i < n; i++){
		int j = sa[Rank[i] - 1];
		if(h > 0) h--;
		for(;j + h < n && i + h < n; h++){
			if(s[j + h] != s[i + h]) break;
		}
		height[Rank[i] - 1] = h;
	} // height sa[i]於sa[i + 1] 的lcp

	// RMQ
	for(int i = 1; i <= n; i++) MIN[i][0] = height[i];
	for(int j = 1; (1 << j) <= n; j++){
		for(int i = 1; i <= n; i++){
			MIN[i][j] = min(MIN[i][j - 1], MIN[i + (1 << (j - 1))][j - 1]);
		}
	}
}
int LCP(int L, int R){ // 0 - (n - 1)
	L = Rank[L]; R = Rank[R]; 
	if(L > R) swap(L, R);
	R--;
	int k = 0; 
	while((1 << (k + 1)) <= R - L + 1) k++; // 2^k可以完全覆蓋
	return min(MIN[L][k], MIN[R - (1 << k) + 1][k]);
}
char A[N], B[N];
int main(){
	int T; scanf("%d", &T); getchar();
	while(T--){
		gets(A); gets(B);
		int l1 = strlen(A); int l2 = strlen(B);
		A[l1] ='$'; for(int i = 0; i <= l2; i++) A[l1 + 1 + i] = B[i];
		string s = (string)A;
		SA(s);
		int ans = 0;
		for(int i = 0; i < n; i++){
			if((sa[i] < l1) != (sa[i + 1] < l1)) {
				ans = max(ans, height[i]);
			}
		}
		printf("Nejdelsi spolecny retezec ma delku %d.\n",ans);
	}
	return 0;
}