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P4081 [USACO17DEC] Standing Out from the Herd P

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P4081 [USACO17DEC] Standing Out from the Herd P

题面

题目链接

解法

首先构建后缀数组。仍然套路地枚举大串的位置$sa[i]$。首先考虑的是不能在别的小串中出现，记录其前后第一个在不同串中出现的后缀位置，分别为$pre_i$和$nxt_i$，则以$sa[i]$开头的后缀对答案贡献的最小长度为$\max(lcs(pre_i, i), lcs(i, nxt_i))$；接下来考虑只在当前小串中出现一次是，则对答案贡献的最小长度为$\max(lcs(i,i+1))$。两者取$\max$即可。总的时间复杂度为$O(n \log n)$。

AC代码

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/**
 * @file:           P4081.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4081
*/
// #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 int long long
#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 = 2e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct SuffixArray {
    int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN], he[MAXN];
    void radix_sort(int m) {
        static int buc[MAXN];
        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, id[MAXN], ed[MAXN], len[MAXN], ans[MAXN];
char buf[MAXN], str[MAXN];
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);
    int pre = 0, mn = INF;
    for (int i = 1; i <= m; ++i) {
        if (id[sa.sa[i]] != id[sa.sa[i - 1]]) {
            int p = id[sa.sa[i - 1]];
            pre = i - 1, mn = ed[p] - sa.sa[i - 1];
        }
        mn = min(mn, sa.he[i]);
        int p = id[sa.sa[i]];
        int cur = max(0ll, ed[p] - sa.sa[i] - max(mn, sa.he[i + 1]));
        ans[p] += cur;
    }
    for (int i = 1; i <= n; ++i)
        write(ans[i]), putchar('\n');
    return 0;
}