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P5028 Annihilate

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P5028 Annihilate

题面

题目链接

解法

题意很简洁，让求$n$个串两两之间的LCS。

既然不能$O(len^2)$的$dp$，那么就使用后缀数组吧。

注意到可以套路地将串拼接，直接枚举大串位置$i$，记录下每个小串的最近的上一个字符位置$pre_j$，用LCS更新答案，$O(n\times len)$。

注意到内存限制为$64M$，无法使用ST表，于是在更新$pre$数组时记录另一个$mn$数组保存从$pre$到$i$的LCS，空间复杂度变成$O(n^2 + len)$，可以通过。

注意到最后的$40pts$其实有考察选手的细心程度，LCS不能大于串的长度。

AC代码

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/**
 * @file:           P5028.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P5028
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 55;
const int MAXM = 1.1e6 + 10;
const int INF = 0x3f3f3f3f;
struct SuffixArray {
    int n, sa[MAXM], rk[MAXM], tp[MAXM], ht[MAXM], he[MAXM];
    void radix_sort(int m) {
        static int buc[MAXM];
        for (int i = 0; i <= m; ++i)
            buc[i] = 0;
        for (int i = 1; i <= n; ++i)
            buc[rk[i]]++;
        for (int i = 1; i <= m; ++i)
            buc[i] += buc[i - 1];
        for (int i = n; i >= 1; --i)
            sa[buc[rk[tp[i]]]--] = tp[i];
    }
    void init(int n, char* s) {
        this->n = n;
        int m = 200;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i] + 1, tp[i] = i;
        radix_sort(m);
        for (int w = 1, p = 0; p < n; m = p, w <<= 1) {
            p = 0;
            for (int i = 1; i <= w; ++i)
                tp[++p] = n - w + i;
            for (int i = 1; i <= n; ++i)
                if (sa[i] > w)
                    tp[++p] = sa[i] - w;
            radix_sort(m);
            copy(rk + 1, rk + n + 1, tp + 1);
            rk[sa[1]] = p = 1;
            for (int i = 2; i <= n; ++i) {
                if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
                    rk[sa[i]] = p;
                else
                    rk[sa[i]] = ++p;
            }
        }
        for (int i = 1, k = 0; i <= n; ++i) {
            if (k)
                k--;
            while (s[i + k] == s[sa[rk[i] - 1] + k])
                k++;
            ht[i] = k;
        }
        for (int i = 1; i <= n; ++i)
            he[i] = ht[sa[i]];
    }
} sa;
int n, m;
int mn[MAXN], ed[MAXN], len[MAXN], pre[MAXN];
int id[MAXM], ans[MAXN][MAXN];
char buf[MAXM], str[MAXM];
signed main() {
    read(n), m = 1;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", buf);
        len[i] = strlen(buf);
        copy(buf, buf + len[i], str + m);
        str[m + len[i]] = '$';
        fill(id + m, id + m + len[i], i);
        ed[i] = m + len[i];
        m += len[i] + 1;
    }
    sa.init(m, str);
    for (int i = 1; i <= m; ++i) {
        for (int j = 1; j <= n; ++j)
            mn[j] = min(mn[j], sa.he[i]);
        int p = id[sa.sa[i]];
        int clen = ed[p] - sa.sa[i];
        pre[p] = i, mn[p] = clen;
        for (int j = 1; j <= n; ++j) {
            if (pre[j]) {
                ans[j][p] = max(ans[j][p], min(mn[j], clen));
                ans[p][j] = max(ans[p][j], min(mn[j], clen));
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (i == j)
                continue;
            write(ans[i][j]), putchar(' ');
        }
        putchar('\n');
    }
    return 0;
}