洛谷 P4248: bzoj 3238: [AHOI2013]差异

题目传送门：洛谷 P4248。

题意简述：

定义两个字符串 (S) 和 (T) 的差异 (operatorname{diff}(S,T)) 为这两个串的长度之和减去两倍的这两个串的最长公共前缀的长度。

给定一个字符串，定义从第 (i) 个字符开始的后缀为 (Suf_i)。

求 (sum_{1le i<jle n}operatorname{diff}(Suf_i,Suf_j))。

题解：

化简式子，原式等于

[egin{align*}&left(sum_{1le i<jle n}i+j ight)-2 imessum_{1le i<jle n}operatorname{lcp}(Suf_i,Suf_j)\=& frac{n(n-1)(n+1)}{2}-2 imessum_{1le i<jle n}operatorname{lcp}(Suf_i,Suf_j)end{align*}]

所以只要求出后半部分即可。

建立字符串的后缀数组。

考虑 Height 数组的贡献：Height 数组中 [2, n] 内的每一个区间都给答案贡献区间最小值。

套路：每个区间的区间最小值之和，使用单调栈解决。

#include <cstdio>
#include <cstring>

typedef long long LL;
const int MN = 500005;

int N;
char str[MN];

int M;
int rk[MN], rk2[MN], SA[MN], SA2[MN];
int buk[MN], cnt;
int Height[MN];

void GetHeight() {
    int k = 0;
    for (int i = 1; i <= N; ++i) {
        if (rk[i] == 1) { k = Height[1] = 0; continue; }
        if (k) --k;
        int j = SA[rk[i] - 1];
        while (i + k <= N && j + k <= N && str[i + k] == str[j + k]) ++k;
        Height[rk[i]] = k;
    }
}

void Rsort() {
    for (int i = 1; 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[SA2[i]]]--] = SA2[i];
}

void GetSA() {
    M = 26;
    for (int i = 1; i <= N; ++i) rk[i] = str[i] - 'a' + 1, SA2[i] = i;
    Rsort();
    for (int j = 1; j < N; j <<= 1) {
        int P = 0;
        for (int i = N - j + 1; i <= N; ++i) SA2[++P] = i;
        for (int i = 1; i <= N; ++i) if (SA[i] > j) SA2[++P] = SA[i] - j;
        Rsort();
        rk2[SA[1]] = P = 1;
        for (int i = 2; i <= N; ++i) {
            if (rk[SA[i]] != rk[SA[i - 1]] || rk[SA[i] + j] != rk[SA[i - 1] + j]) ++P;
            rk2[SA[i]] = P;
        }
        for (int i = 1; i <= N; ++i) rk[i] = rk2[i];
        M = P;
        if (M == N) break;
    }
    GetHeight();
}

int st[MN], t;
int L[MN], R[MN];

int main() {
    scanf("%s", str + 1);
    N = strlen(str + 1);
    GetSA();
    st[t = 1] = 1;
    for (int i = 2; i <= N; ++i) {
        while (t && Height[st[t]] > Height[i]) R[st[t--]] = i;
        L[i] = st[t];
        st[++t] = i;
    } while (t) R[st[t--]] = N + 1;
    LL Ans = (LL)(N - 1) * N * (N + 1) / 2;
    for (int i = 2; i <= N; ++i)
        Ans -= 2ll * (R[i] - i) * (i - L[i]) * Height[i];
    printf("%lld
", Ans);
    return 0;
}