Codeforces 666E E

E - Forensic Examination

我也不知道为什么这个复杂度能过， 而且跑得还挺快， 数据比较水？

在sa上二分出上下界， 然后莫队 + 线段树维护区间众数。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 6e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = acos(-1);

#define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb))
#define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2)

int r[3 * N], sa[3 * N];
int wa[3 * N], wb[3 * N], wv[3 * N], ss[3 * N];
int rnk[N], lcp[N];

int c0(int *r, int a, int b) {return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2]; }

int c12(int k, int *r, int a, int b) {
    if(k == 2) return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
    return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}

void sort(int *r, int *a, int *b, int n, int m) {
    for(int i = 0; i < n; i++) wv[i] = r[a[i]];
    for(int i = 0; i < m; i++) ss[i] = 0;
    for(int i = 0; i < n; i++) ss[wv[i]]++;
    for(int i = 1; i < m; i++) ss[i] += ss[i - 1];
    for(int i = n - 1; i >= 0; i--) b[--ss[wv[i]]] = a[i];
}

void DC3(int *r, int *sa, int n, int m) {
    int i, j, p;
    int *san = sa + n, *rn = r + n;
    int ta = 0, tb = (n + 1) / 3, tbc = 0;

    r[n] = r[n + 1] = 0;
    for(i = 0; i < n; i++) if(i % 3 != 0) wa[tbc++] = i;

    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);

    for(p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
        rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++;

    if(p < tbc) DC3(rn, san, tbc, p);
    else for(i = 0; i < tbc; i++) san[rn[i]] = i;

    for(i = 0; i < tbc; i++) if(san[i] < tb) wb[ta++] = san[i] * 3;
    if(n % 3 == 1) wb[ta++] = n - 1;

    sort(r, wb, wa, ta, m);

    for(i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i;
    for(i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
        sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];

    for( ; i < ta; p++) sa[p] = wa[i++];
    for( ; j < tbc; p++) sa[p] = wb[j++];
}


void calc_LCP(int *r, int *sa, int n) {
    int k = 0;
    for(int i = 0; i < n; i++) rnk[sa[i]] = i;
    for(int i = 0; i < n - 1; lcp[rnk[i++]] = k) {
        if(k) k--;
        for(int j = sa[rnk[i] - 1]; r[i + k] == r[j + k]; k++);
    }
}

int Log[N];
struct ST {
    int dp[N][20], ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline int query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
};

namespace SGT {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    const int N = 5e4 + 7;
    PII a[N << 2];
    int b[N];
    void build(int l, int r, int rt) {
        if(l == r) {
            a[rt].se = -l;
            b[l] = rt;
            return;
        }
        int mid = (l + r) >> 1;
        build(lson); build(rson);
        a[rt] = max(a[rt << 1], a[rt << 1 | 1]);
    }
    inline void update(int x, int val) {
        x = b[x];
        a[x].fi += val;
        x >>= 1;
        for( ; x; x >>= 1)
            a[x] = max(a[x << 1], a[x << 1 | 1]);
    }
    PII query(int L, int R, int l, int r, int rt) {
        if(R < l || r < L) return mk(-inf, inf);
        if(L <= l && r <= R) return a[rt];
        int mid = (l + r) >> 1;
        return max(query(L, R, lson), query(L, R, rson));
    }
}

struct Qus {
    int L, R, numl, numr, id;
    bool operator < (const Qus& rhs) const {
        if(L / 750 == rhs.L / 750) return R < rhs.R;
        return L < rhs.L;
    }
};

int n, m, q, mxc = 256, who[N];
PII ans[N];
char s[N];
ST rmq;
Qus qus[N];

int main() {
    scanf("%s", s);
    for(int i = 0; s[i]; i++)
        r[n++] = s[i], mxc = max(mxc, (int)s[i]);
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        r[n++] = mxc++;
        scanf("%s", s);
        for(int j = 0; s[j]; j++)
            r[n] = s[j], who[n++] = i;
    }
    r[n] = 0;
    DC3(r, sa, n + 1, mxc);
    calc_LCP(r, sa, n + 1);
    rmq.build(n, lcp, -1);
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        int l, r, pl, pr;
        scanf("%d%d%d%d", &l, &r, &pl, &pr);
        int pos = rnk[pl - 1], len = pr - pl + 1;
        int low = 1, high = pos - 1, L = pos, R = pos;
        while(low <= high) {
            int mid = (low + high) >> 1;
            if(rmq.query(mid + 1, pos) >= len) L = mid, high = mid - 1;
            else low = mid + 1;
        }
        low = pos + 1, high = n;
        while(low <= high) {
            int mid = (low + high) >> 1;
            if(rmq.query(pos + 1, mid) >= len) R = mid, low = mid + 1;
            else high = mid - 1;
        }
        qus[i] = Qus{L, R, l, r, i};
    }
    SGT::build(1, m, 1);
    sort(qus + 1, qus + 1 + q);
    int l = 1, r = 0;
    for(int i = 1; i <= q; i++) {
        int L = qus[i].L, R = qus[i].R, numl = qus[i].numl, numr = qus[i].numr;
        while(r < R) r++, SGT::update(who[sa[r]], 1);
        while(l > L) l--, SGT::update(who[sa[l]], 1);
        while(r > R) SGT::update(who[sa[r]], -1), r--;
        while(l < L) SGT::update(who[sa[l]], -1), l++;
        ans[qus[i].id] = SGT::query(numl, numr, 1, m, 1);
    }
    for(int i = 1; i <= q; i++) printf("%d %d
", -ans[i].se, ans[i].fi);
    return 0;
}

/*
*/