【后缀数组】【SP1811】 LCS - Longest Common Substring

题目链接

题意翻译

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 。

思路

求两个串的最长公共子串

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 500010
using namespace std;

int n, n1, n2;

char s1[maxn], s2[maxn], s[maxn]; 

int tax[maxn], rk[maxn], tp[maxn], sa[maxn], M = 200;
void rsort() {
    for (int i = 0; i <= M; ++i) tax[i] = 0;
    for (int i = 1; i <= n; ++i) ++tax[rk[i]];
    for (int i = 1; i <= M; ++i) tax[i] += tax[i - 1];
    for (int i = n; i; --i) sa[tax[rk[tp[i]]]--] = tp[i];
}

int H[maxn];
void SA() {
    for (int i = 1; i <= n; ++i) rk[i] = s[i], tp[i] = i;
    int c1 = 0; rsort();
    for (int k = 1; k <= n; k *= 2) {
        if (c1 == n) break; M = c1; c1 = 0;
        for (int i = n - k + 1; i <= n; ++i) tp[++c1] = i;
        for (int i = 1; i <= n; ++i) if (sa[i] > k) tp[++c1] = sa[i] - k;
        rsort(); swap(tp, rk); rk[sa[1]] = c1 = 1;
        for (int i = 2; i <= n; ++i) {
            if (tp[sa[i - 1]] != tp[sa[i]] || tp[sa[i - 1] + k] != tp[sa[i] + k]) ++c1;
            rk[sa[i]] = c1;
        }
    }
    int lcp = 0;
    for (int i = 1; i <= n; ++i) {
        if (lcp) --lcp;
        int j = sa[rk[i] - 1];
        while (s[j + lcp] == s[i + lcp]) ++lcp;
        H[rk[i]] = lcp;
    }
}

int ans;
int main() {
    scanf("%s%s", s1 + 1, s2 + 1);
    n = n1 = strlen(s1 + 1); n2 = strlen(s2 + 1);
    for (int i = 1; i <= n1; ++i) s[i] = s1[i];
    s[++n] = '$';
    for (int i = 1; i <= n2; ++i) s[++n] = s2[i]; SA();
    for (int i = 2; i <= n; ++i) {
        int x = sa[i], y = sa[i - 1];
        if (x > y) swap(x, y);
        if (x <= n1 && y > n1 + 1) ans = max(ans, H[i]);
    } cout << ans << endl;
    return 0;
}