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  • Codeforces 204E Little Elephant and Strings 后缀数组 + 并查集

    按lcp合并计算答案。

    #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 = 2e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    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;
         }
    }
    
    int fa[N], dl[N], dr[N];
    
    int n, k, maxH, belong[N], to[N], cnt[N];
    LL ans[N], tmp[N];
    char s[N];
    
    vector<int> V[N];
    vector<int> ID;
    
    int getRoot(int x) {
        return x == fa[x] ? x : fa[x] = getRoot(fa[x]);
    }
    
    int Merge(int u, int v) {
        int x = getRoot(u);
        int y = getRoot(v);
        fa[y] = x;
        chkmin(dl[x], dl[y]);
        chkmax(dr[x], dr[y]);
        return x;
    }
    
    void printSuf(int x) {
        for(int i = sa[x]; i <= san; i++) putchar((char)r[i]);
        for(int i = 0; i < (sa[x] + 3); i++) putchar(' ');
        printf("sa: %d  lcp: %d
    ", sa[x], lcp[x]);
    }
    
    int main() {
        scanf("%d%d", &n, &k);
        for(int i = 1; i <= n; i++) {
            if(i > 1) r[san++] = maxc++;
            scanf("%s", s);
            int len = strlen(s);
            for(int j = 0; s[j]; j++) {
                r[san] = s[j];
                belong[san] = i;
                san++;
            }
            if(k == 1) ans[i] = 1LL * len * (len + 1) / 2;
        }
    
        if(k == 1) {
            for(int i = 1; i <= n; i++) {
                printf("%lld ", ans[i]);
            }
            puts("");
            return 0;
        }
    
        r[san] = 0;
        buildSa(r, san + 1, maxc);
    
        for(int i = 1; i <= san; i++) {
            V[lcp[i]].push_back(i);
            chkmax(maxH, lcp[i]);
        }
    
        int now = 0;
    
        for(int i = 1, j = 1; i <= san; i++) {
            j = max(j, i);
            while(now < k && j <= san) {
                if(belong[sa[j]]) {
                    if(!cnt[belong[sa[j]]]) now++;
                    cnt[belong[sa[j]]]++;
                }
                j++;
            }
            if(now < k) break;
            to[i] = j - 1;
            if(belong[sa[i]]) {
                if(cnt[belong[sa[i]]] == 1) now--;
                cnt[belong[sa[i]]]--;
            }
        }
    
        for(int i = 1; i <= san; i++) {
            fa[i] = dl[i] = dr[i] = i;
        }
    
        for(int i = maxH; i > 0; i--) {
            ID.clear();
            for(auto &t : V[i]) {
                ID.push_back(Merge(t - 1, t));
            }
            for(auto &t : ID) {
                t = getRoot(t);
            }
            sort(ALL(ID));
            ID.erase(unique(ALL(ID)), ID.end());
            for(auto &t : ID) {
                if(!to[dl[t]] || dr[t] < to[dl[t]]) continue;
                int x = lcp[dl[t]];
                if(dr[t] + 1 <= san) chkmax(x, lcp[dr[t] + 1]);
                tmp[dl[t]] += i - x;
                tmp[dr[t] + 1] -= i - x;
            }
        }
    
        for(int i = 1; i <= san; i++) {
            tmp[i] += tmp[i - 1];
            ans[belong[sa[i]]] += tmp[i];
        }
    
        for(int i = 1; i <= n; i++) {
            printf("%lld ", ans[i]);
        }
        puts("");
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/11132851.html
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