题面

大秦为你打开题目传送门

题目描述

给定两个字符串，请找出它们的最长公共子串。

思路

由于本人不会后缀数组，所以我们就哈希淼过罢。

直接大力二分哈希找长度，搞定。

代码

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#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;
typedef unsigned long long ull;

namespace FastIO 
{
    // char buf[1 << 20], *p1, *p2;
    // #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
};
using namespace FastIO;

const int N = 1e5 + 10;
const int B = 31;

char s[N], t[N];
int ls, lt;

ull Power[N];
ull hshS[N], hshT[N];
inline ull Hash(ull* a, int l, int r) { return a[r] - a[l - 1] * Power[r - l + 1]; }

inline void Input() {
    scanf("%s%s", s + 1, t + 1);
}

inline void Init() {
    ls = strlen(s + 1);
    lt = strlen(t + 1);
    Power[0] = 1;
    for(int i = 1; s[i]; i++) {
        hshS[i] = hshS[i - 1] * B + (s[i] - 'a');
        Power[i] = Power[i - 1] * B;
    }
    for(int i = 1; t[i]; i++) {
        hshT[i] = hshT[i - 1] * B + (t[i] - 'a');
        Power[i] = Power[i - 1] * B;
    }
}

inline bool Check(int len) {
    vector<ull >v;
    for(int i = 1; i + len - 1 <= ls; i++) {
        v.push_back(Hash(hshS, i, i + len - 1));
    }
    sort(v.begin(), v.end());
    for(int i = 1; i + len - 1 <= lt; i++) {
        int pos = lower_bound(v.begin(), v.end(), Hash(hshT, i, i + len - 1)) - v.begin();
        if(v[pos] == Hash(hshT, i, i + len - 1)) {
            return true;
        }
    }
    return false;
}

inline void Work() {
    Init();
    int l = 0, r = min(ls, lt), best = -1;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(Check(mid)) { // This len is right
            l = mid + 1;
            best = mid;
        }
        else {
            r = mid - 1;
        }
    }
    printf("%d", best);
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}