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  • Codeforces 547E Mike and Friends 后缀数组

    Mike and Friends

    求出后缀数组, 对于每个询问二分出左右端点, 离线之后用树状数组求就好了。

    或者不建st表, 用并查集求出左右端点, 这样空间和常数都更优。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned 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 ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 5e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}
    
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    
    int r[N], sa[N], _t[N], _t2[N], c[N], rk[N], lcp[N], san;
    int maxc = 'z' + 1;
    
    void buildSa(int *r, int n, int m) {
        int i, j = 0, k = 0, *x = _t, *y = _t2;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[i] = r[i]]++;
        for(i = 1; i < m; i++) c[i] += c[i - 1];
        for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
        for(int k = 1; k <= n; k <<= 1) {
            int p = 0;
            for(i = n - k; i < n; i++) y[p++] = i;
            for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
            for(i = 0; i < m; i++) c[i] = 0;
            for(i = 0; i < n; i++) c[x[y[i]]]++;
            for(i = 1; i < m; i++) c[i] += c[i - 1];
            for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
            swap(x, y);
            p = 1; x[sa[0]] = 0;
            for(int i = 1; i < n; i++) {
                if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
                    x[sa[i]] = p - 1;
                else x[sa[i]] = p++;
            }
            if(p >= n) break;
            m = p;
         }
         for(i = 1; i < n; i++) rk[sa[i]] = i;
         for(i = 0; i < n - 1; i++) {
            if(k) k--;
            j = sa[rk[i] - 1];
            while(r[i + k] == r[j + k]) k++;
            lcp[rk[i]] = k;
         }
    }
    
    struct Bit {
        int a[N];
        void modify(int x, int v) {
            for(int i = x; i < N; i += i & -i) {
                a[i] += v;
            }
        }
        int sum(int x) {
            int ans = 0;
            for(int i = x; i; i -= i & -i) {
                ans += a[i];
            }
            return ans;
        }
        int query(int L, int R) {
            if(L > R) return 0;
            return sum(R) - sum(L - 1);
        }
    } bit;
    
    int fa[N], dl[N], dr[N];
    
    int getRoot(int x) {
        return fa[x] == x ? x : fa[x] = getRoot(fa[x]);
    }
    
    void Merge(int u, int v) {
        int x = getRoot(u);
        int y = getRoot(v);
        if(x == y) return;
        chkmin(dl[x], dl[y]);
        chkmax(dr[x], dr[y]);
        fa[y] = x;
    }
    
    struct Qus {
        int vl, vr, id;
    };
    
    int n, q, belong[N], len[N], sid[N];
    int ans[N];
    char s[N];
    vector<Qus> Q[N];
    
    int id[N], L[N], R[N], lcpId[N];
    
    void printSuf(int x) {
        for(int i = sa[x]; i < san; i++) putchar((char)r[i]);
        for(int i = 0; i < (sa[x] + 5); i++) putchar(' ');
        printf("i: %3d  sa: %3d  lcp: %3d  belong: %3d
    ", x, sa[x], lcp[x], belong[sa[x]]);
    }
    
    
    int main() {
        scanf("%d%d", &n, &q);
        for(int i = 1; i <= n; i++) {
            if(i > 1) r[san++] = maxc++;
            scanf("%s", s);
            len[i] = strlen(s);
            sid[i] = san;
            for(int j = 0; s[j]; j++) {
                r[san] = s[j];
                belong[san++] = i;
            }
            id[i] = i;
        }
        r[san] = 0;
        buildSa(r, san + 1, maxc);
    
        for(int i = 1; i <= san; i++) {
            fa[i] = i;
            dl[i] = dr[i] = i;
            lcpId[i] = i;
        }
    
        sort(lcpId + 1, lcpId + san + 1, [&](int x, int y) {
            return lcp[x] > lcp[y];
        });
    
        sort(id + 1, id + 1 + n, [&](int x, int y) {
            return len[x] > len[y];
        });
    
        for(int i = 1, j = 1; i <= n; i++) {
            while(j <= san && lcp[lcpId[j]] >= len[id[i]]) {
                Merge(lcpId[j], lcpId[j] - 1);
                j++;
            }
            int Rt = getRoot(rk[sid[id[i]]]);
            L[id[i]] = dl[Rt];
            R[id[i]] = dr[Rt];
        }
    //    puts("");
    //    for(int i = 1; i <= san; i++) {
    //        printSuf(i);
    //    }
        for(int i = 1; i <= q; i++) {
            int l, r, k;
            scanf("%d%d%d", &l, &r, &k);
            Q[L[k] - 1].push_back(Qus{l, r, -i});
            Q[R[k]].push_back(Qus{l, r, i});
        }
        for(int i = 1; i <= san; i++) {
            if(belong[sa[i]]) {
                bit.modify(belong[sa[i]], 1);
            }
            for(auto &q : Q[i]) {
                if(q.id > 0) ans[q.id] += bit.query(q.vl, q.vr);
                else ans[-q.id] -= bit.query(q.vl, q.vr);
            }
        }
        for(int i = 1; i <= q; i++) {
            printf("%d
    ", ans[i]);
        }
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/11180651.html
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