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P3181 [HAOI2016] 找相同字符串

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P3181 [HAOI2016] 找相同字符串

题面

题目链接

解法

假设我们已经知道如何计算单一串的答案，之后可以将两个串拼接，中间间隔字符$，算出新串的答案。但是这样会重复计算两个子串在同一个原串，所以还需减去两个原串的答案。即$ans=anss-ans{s1}-ans_{s2}$。

现在只需考虑如何算出一个单一串的答案。根据后缀数组的性质，显然答案为

算出每个$height$对$ans$的贡献$(i - pre_i) \times (nxt_i - i) \times height(i)$，其中$pre_i$表示前一个小于$height(i)$的位置，$nxt_i$表示后一个小于等于$height_i$的位置。单调栈维护。

时间复杂度$O(n \log n)$，但显然可以通过使用更高级的$SA$优化到$O(n)$。

AC代码

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/**
 * @file:           P3181.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P3181
*/
// #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 = 4e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct suffix_array {
    char *str;
    int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN];
    void radix_sort(int m) {
        static int buk[MAXN];
        for (int i = 0; i <= m; ++i)
            buk[i] = 0;
        for (int i = 1; i <= n; ++i)
            buk[rk[i]]++;
        for (int i = 1; i <= m; ++i)
            buk[i] += buk[i - 1];
        for (int i = n; i >= 1; --i)
            sa[buk[rk[tp[i]]]--] = tp[i];
    }
    void build(char *s, int n) {
        this->str = s;
        this->n = n;
        int m = 128;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i] + 1, tp[i] = i;
        radix_sort(m);
        for (int p = 0, w = 1; 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;
        }
    }
    int calc() {
        int ret = 0, top;
        static int pre[MAXN], nxt[MAXN], sta[MAXN];
        sta[top = 0] = 0;
        for (int i = 1; i <= n; ++i) {
            while (top && ht[sa[sta[top]]] >= ht[sa[i]])
                --top;
            pre[i] = i - sta[top];
            sta[++top] = i;
        }
        sta[top = 0] = n + 1;
        for (int i = n; i >= 1; --i) {
            while (top && ht[sa[sta[top]]] > ht[sa[i]])
                --top;
            nxt[i] = sta[top] - i;
            sta[++top] = i;
        }
        for (int i = 1; i <= n; ++i)
            ret += pre[i] * nxt[i] * ht[sa[i]];
        return ret;
    }
};
int n0, n1, n2;
char s0[MAXN], s1[MAXN], s2[MAXN];
suffix_array sa0, sa1, sa2;
signed main() {
    scanf("%s%s", s1 + 1, s2 + 1);
    n1 = strlen(s1 + 1);
    n2 = strlen(s2 + 1);
    copy(s1 + 1, s1 + n1 + 1, s0 + 1);
    copy(s2 + 1, s2 + n2 + 1, s0 + n1 + 2);
    s0[n1 + 1] = '$', n0 = n1 + n2 + 1;
    sa0.build(s0, n0);
    sa1.build(s1, n1);
    sa2.build(s2, n2);
    int ans = sa0.calc() - sa1.calc() - sa2.calc();
    write(ans), putchar('\n');
    return 0;
}