放爆竹简略题解

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1.前言

虽然猜对了结论，但还是要 yy 一下证明。

变态，卡 STL

2.题解

我们只需要一个结论：对于两个字符串 $A,B(|A|\le |B|)$，它们的答案不超过 $2|B|$

反证法：

记 $n=|A|$

如图，绿色方框表示 $A$，红色方框表示 $B$。

由图可以看出：若答案超过了 $2|B|$，则可以得出一个必要条件：$\exists x$，使得 $A$ 的前 $x$ 和 $A$ 的后 $x$ 相匹配， $A$ 的后 $n-x$ 和 $A$ 的前 $n-x$ 相匹配。

如图，两个绿色方框表示字符串$A$。

可以推出，红色部分与蓝色蓝色部分相匹配。

继续推导，可以发现 $A$ 的循环节为 $x$，且 $\frac{|A|}{x}\in Z^{\ast }$。则 $A$ 是 $B$ 的循环节，与题意不符。
$\square$

用 $Hash$ ，枚举 $A$ 串的长度，统计在当前枚举的 $B$ 串中最大匹配数量，用 unordered_map 保存

($pts$ 60) 变态，卡 STL

#include <map>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

#define LL long long
#define ULL unsigned long long
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T>
void write (T x) {
    if (x < 0) {
        putchar ('-');
        x = -x;
    }
    if (x < 10) {
        putchar (x + '0');
        return;
    }
    write (x / 10);
    putchar (x % 10 + '0');
}
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

const int Maxn = 2 * 1e4;
const int Maxlen = 500;
const int P = 31;

int n;
char s[Maxlen * 2 + 5];
string c[Maxn + 5];
ULL Hash[Maxlen * 2 + 5], w[Maxlen * 2 + 5];
unordered_map <ULL, int> mp;

ULL Query (int l, int r) {
    return Hash[r] - Hash[l - 1] * w[r - l + 1];
}
bool cmp (string x, string y) {
    return x.length () > y.length ();
}

int main () {
    w[0] = 1;
    for (int i = 1; i <= Maxlen * 2; i++) w[i] = w[i - 1] * P;
    
    int ans = 0;
    read (n);
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }
    sort (c + 1, c + 1 + n, cmp);
    //从大到小排序，保证靠前的字符串是 B
    for (int k = 1; k <= n; k++) {
        memset (s, 0, sizeof s);
        for (int i = 0; i < c[k].length (); i++) {
            s[i + 1] = c[k][i];
        }
        //s为 B串
        
        int len = strlen (s + 1);
        for (int i = 1; i <= len; i++) {
            s[i + len] = s[i];
        }
        for (int i = 1; i <= len * 2; i++) {
            Hash[i] = Hash[i - 1] * P + s[i] - '0' + 1;
        }
        ULL idx = 0;
        for (int i = 1; i <= len; i++) {
            idx = idx * P + s[i] - '0' + 1;
        }
        if (mp.find (idx) != mp.end ())
            ans = Max (ans, mp[idx]);
        for (int i = 1; i <= len; i++) {//枚举|A|
            int res = i;
            for (int j = i + 1; j + i - 1 <= len * 2; j += i) {
                if (Query (j - i, j - 1) == Query (j, j + i - 1)) res += i;
                else break;
            }
            for (int p = res + 1, q = 1; p <= len * 2; p++, q++) {
                if (s[p] == s[q]) res++;
                else break;
            }
            
            ULL idx = 0;
            for (int j = 1; j <= i; j++) {
                idx = idx * P + s[j] - '0' + 1;
            }
            //当A 串为idx 时，答案为 res
            if (mp.find (idx) == mp.end ()) mp[idx] = res;
            else mp[idx] = Max (mp[idx], res);
            //记录下答案
        }
    }
    write (ans);
    return 0;
}

用 $Tire$ 树维护

#include <map>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

#define LL long long
#define ULL unsigned long long
template <typename T>
void read (T &x) {
    x = 0; T f = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar ();
    }
    x *= f;
}
template <typename T>
void write (T x) {
    if (x < 0) {
        putchar ('-');
        x = -x;
    }
    if (x < 10) {
        putchar (x + '0');
        return;
    }
    write (x / 10);
    putchar (x % 10 + '0');
}
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

const int Maxn = 2 * 1e4;
const int Maxlen = 500;

int n;
string c[Maxn + 5];

bool cmp (string x, string y) {
    return x.length () > y.length ();
}

int tot;
int Tire[2][Maxn * Maxlen * 2 + 5];
void Update (string s) {
    s += s;
    int p = 0;
    for (int i = 0; i < s.length (); i++) {
        int ch = s[i] - '0';
        if (Tire[ch][p] == 0)
            Tire[ch][p] = ++tot;
        p = Tire[ch][p];
    }
}
int Query (string s) {
    int p = 0, res = 0;
    for (int i = 0; ; i = (i + 1) % (int)s.length ()) {
        int ch = s[i] - '0';
        if (Tire[ch][p] == 0) return res;
        res++; p = Tire[ch][p];
    }
    return res;
}

int main () {
    read (n);
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }
    sort (c + 1, c + 1 + n, cmp);
    int ans = 0;
    for (int k = 1; k <= n; k++) {
        ans = Max (ans, Query (c[k]));
        Update (c[k]);
    }
    write (ans);
    return 0;
}