NOI2016優秀的拆分

• 一種想法是枚舉分割位置， 而後考慮前面部分有多少種可行的AA拆分方式， 後面部分有多少種可行的BB拆分方式， 而後乘法原理便可
• 那麼問題是如何快速求出合法方案
• 解法是首先枚舉長度len， 而後將序列分紅$\frac{n}{len}$段， 而後咱們對於每一個連續的三個段i，j, k， 求一下i, j的最長公共後綴a， j，k的最長公共前綴b,
• 若是a + b < len， 顯然沒法構造
• 若是a + b >= len， 那麼整整一個區間均可以構造出來， 差分統計一下答案
• 由於某些地方沒清空以及某些地方特判寫狗了， WA成狗

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 60010
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
int log[M];
struct SA {
    char s[M];
    int sa[M], tmp[M], tex[M], rnk[M], f[M][17], n, m, q;

    bool check(int i, int j, int w) {
        return tmp[i] == tmp[j] && tmp[i + w] == tmp[j + w];
    }

    void qsort() {
        for(int i = 0; i <= m; i++) tex[i] = 0;
        for(int i = 1; i <= n; i++) tex[rnk[tmp[i]]]++;
        for(int i = 1; i <= m; i++) tex[i] += tex[i - 1];
        for(int i = n; i >= 1; i--) sa[tex[rnk[tmp[i]]]--] = tmp[i];
    }

    void suffix() {
        memset(tmp, 0, sizeof(tmp));
        memset(tex, 0, sizeof(tex));
        memset(sa, 0, sizeof(sa));
        memset(rnk, 0, sizeof(rnk));
        memset(f, 0, sizeof(f));
        m = 127, q = 0;
        for(int i = 1; i <= n; i++) rnk[i] = s[i], tmp[i] = i;
        qsort();
        for(int w = 1; q < n; w <<= 1, m = q) {
            q = 0;
            for(int i = n - w + 1; i <= n; i++) tmp[++q] = i;
            for(int i = 1; i <= n; i++) if(sa[i] > w) tmp[++q] = sa[i] - w;
            qsort();
            swap(rnk, tmp);
            rnk[sa[1]] = q = 1;
            for(int i = 2; i <= n; i++) if(check(sa[i - 1], sa[i], w)) rnk[sa[i]] = q;
                else rnk[sa[i]] = ++q;
        }
        q = 0;
    //  cerr << "\n";
        for(int i = 1; i <= n; i++) {
            if(q) q--;
            int j = sa[rnk[i] + 1];
            if(rnk[i] == n) continue;
            while(s[i + q] == s[j + q] && i + q <= n && j + q <= n) q++;
            f[rnk[i]][0] = q;
        //  cerr << q << " ";
            //if(q >n) cerr<<"!";
        }
    //  cerr << "\n";
        //for(int i = 1; i <= n; i++) cerr << rnk[i] << " ";
    //  cout << "\n";
        for(int j = 1; j <= 15; j++) {
            for(int i = 1; i <= n; i++) {
                if(i + (1 << (j - 1)) > n) break;
                f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
            }
        }
    }

    int lcp(int a, int b) {
        a = rnk[a], b = rnk[b];
        if(a > b) swap(a, b);
        int k = log[b - a];
        return min(f[a][k], f[b - (1 << k)][k]);
    }
} a, b;
int be[M], ed[M];
char s[M];

void work() {
    memset(s, 0, sizeof(s));
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    for(int i = 1; i <= n; i++) {
        a.s[i] = s[i], b.s[i] = s[n - i + 1];
    }
    memset(be, 0, sizeof(be));
    memset(ed, 0, sizeof(ed));
    a.n = b.n = n;
    a.suffix();
    b.suffix();
    for(int len = 1; len <= n / 2; len++) {
        for(int i = len, j = i + len; j <= n; i += len, j += len) {
            int x = min(a.lcp(i, j), len), y = min(b.lcp(n - (i - 1) + 1, n - (j - 1) + 1), len - 1);
            int t = x + y - len + 1;
            if(x + y >= len) {
                be[i - y]++;
                be[i - y + t]--;
                ed[j + x - t]++;
                ed[j + x]--;
            }
        }
    }
    ll ans = 0;
    for(int i = 1; i <= n; i++) be[i] += be[i - 1], ed[i] += ed[i - 1], ans += 1ll * ed[i - 1] * be[i];
    cout << ans << "\n";
}

int main() {
    for(int i = 2; i < M; i++) log[i] = log[i >> 1] + 1;
    int T = read();
    while(T--) work();
    return 0;
}


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