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 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;
const double eps = 1e-;
const double PI = acos(-);

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

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

namespace SGT {
    int tot = , Rt[N];
    struct Node {
        int sum, ls, rs;
    } a[N * ];
    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 >> ;
        if(p <= mid) modify(p, l, mid, a[x].ls, a[y].ls);
        else modify(p, mid + , 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 >> ;
        if(p <= mid) query(p, l, mid, a[x].ls);
        else query(p, mid + , 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 = , k = , *x = t, *y = t2;
    for(i = ; i < m; i++) c[i] = ;
    for(i = ; i < n; i++) c[x[i] = s[i]]++;
    for(i = ; i < m; i++) c[i] += c[i - ];
    for(i = n - ; i >= ; i--) sa[--c[x[i]]] = i;
    for(int k = ; k <= n; k <<= ) {
        int p = ;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = ; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = ; i < m; i++) c[i] = ;
        for(i = ; i < n; i++) c[x[y[i]]]++;
        for(i = ; i < m; i++) c[i] += c[i - ];
        for(i = n - ; i >= ; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = ; x[sa[]] = ;
        for(int i = ; i < n; i++) {
            if(y[sa[i - ]] == y[sa[i]] && y[sa[i - ] + k] == y[sa[i] + k])
                x[sa[i]] = p - ;
            else x[sa[i]] = p++;
        }
        if(p >= n) break;
        m = p;
     }

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

/*
*/