Codeforces 653F Paper task SA

Paper task

如果不要求本质不同直接st表二分找出最右端， 然后计数就好了。

要求本质不同， 先求个sa， 然后用lcp求本质不同就好啦。

#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 = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);

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]);
    }
    int query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y - (1 << k) + 1][k]);
    }
};

const int MAX = 6e5;
const int MIN = -6e5;

namespace SGT {
    int tot = 0, Rt[N];
    struct Node {
        int sum, ls, rs;
    } a[N * 25];
    void modify(int p, int l, int r, int& x, int y) {
        x = ++tot; a[x] = a[y]; a[x].sum++;
        if(l == r) return;
        int mid = l + r >> 1;
        if(p <= mid) modify(p, l, mid, a[x].ls, a[y].ls);
        else modify(p, mid + 1, r, a[x].rs, a[y].rs);
    }
    int query(int p, int l, int r, int x) {
        if(l == r) return a[x].sum;
        int mid = l + r >> 1;
        if(p <= mid) query(p, l, mid, a[x].ls);
        else query(p, mid + 1, r, a[x].rs);
    }
}

int sa[N], t[N], t2[N], c[N], rk[N], lcp[N];
void buildSa(char *s, 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] = s[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(s[i + k] == s[j + k]) k++;
        lcp[rk[i]] = k;
     }
}

int n, a[N], p[N];
char s[N];
ST rmq;

int main() {
    scanf("%d", &n);
    scanf("%s", s + 1);
    for(int i = 1; i <= n; i++) {
        a[i] = s[i] == ‘(‘ ? 1 : -1;
        a[i] += a[i - 1];
    }
    rmq.build(n, a, -1);
    for(int i = 1; i <= n; i++)
        SGT::modify(a[i], MIN, MAX, SGT::Rt[i], SGT::Rt[i - 1]);
    for(int i = 1; i <= n; i++) {
        if(s[i] == ‘)‘) continue;
        int low = i + 1, high = n; p[i] = i;
        while(low <= high) {
            int mid = low + high >> 1;
            if(rmq.query(i, mid) >= a[i] - 1) p[i] = mid, low = mid + 1;
            else high = mid - 1;
        }
    }
    LL ans = 0;
    buildSa(s + 1, n + 1, 255);
    for(int i = 1; i <= n; i++) {
        int x = sa[i] + 1;
        if(s[x] == ‘)‘) continue;
        int dn = x + lcp[i], up = p[x];
        if(dn <= up) {
            ans += SGT::query(a[x] - 1, MIN, MAX, SGT::Rt[up]);
            ans -= SGT::query(a[x] - 1, MIN, MAX, SGT::Rt[dn - 1]);
        }
    }
    printf("%lld\n", ans);
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/10648337.html