Codeforces 524F And Yet Another Bracket Sequence 哈希

And Yet Another Bracket Sequence

枚举起点， 增加的(肯定在最前面， 增加的)肯定在最后面， 比字典序用hash， 卡了自然溢出。。

#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 = 2e6 + 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;}

const int B = 23333;
int n;
char s[N];
int a[N];

LL hs[N], powB[N], powL[N], powR[N];

int cntL[N], cntR[N];

int Log[N];
struct ST {
    int dp[N][21], 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]);
    }
} rmq;

inline LL getHash(int L, int R) {
    return ((hs[R] - hs[L - 1] * powB[R - L + 1] % mod) + mod) % mod;
}

inline LL getPosHash(int i, int len) {
    if(len <= cntL[i]) return powL[len];
    if(len <= cntL[i] + n) return (powL[cntL[i]] * powB[len - cntL[i]] % mod + getHash(i, i + len - cntL[i] - 1)) % mod;
    return (powR[len - cntL[i] - n] + getHash(i, i + n - 1) * powB[len - cntL[i] - n] % mod + powL[cntL[i]] * powB[len - cntL[i]] % mod) % mod;
}

inline char getch(int i, int p) {
    if(p <= cntL[i]) return ‘(‘;
    else if(p <= cntL[i] + n) return s[i + p - cntL[i] - 1];
    return ‘)‘;
}

bool check(int p1, int p2, int len) {
    int low = 1, high = len, mid, p = -1;
    while(low <= high) {
        mid = low + high >> 1;
        if(getPosHash(p1, mid) != getPosHash(p2, mid)) p = mid, high = mid - 1;
        else low = mid + 1;
    }
    if(p == -1) return false;
    return getch(p1, p) < getch(p2, p);
}

int main() {
    for(int i = powB[0] = 1; i < N; i++) powB[i] = powB[i - 1] * B % mod;
    for(int i = 1; i < N; i++) powL[i] = (powL[i - 1] * B + ‘(‘) % mod;
    for(int i = 1; i < N; i++) powR[i] = (powR[i - 1] * B + ‘)‘) % mod;
    scanf("%s", s + 1);
    n = strlen(s + 1);
    for(int i = 1; i <= n; i++) s[i + n] = s[i];
    for(int i = 1; i <= 2 * n; i++) hs[i] = (hs[i - 1] * B + s[i]) % mod;
    for(int i = 1; i <= 2 * n; i++) a[i] = a[i - 1] + (s[i] == ‘(‘ ? 1 : -1);
    rmq.build(2 * n, a, -1);
    int need = inf;
    for(int i = 1; i <= n; i++) {
        int mn = rmq.query(i, i + n - 1);
        if(mn < a[i - 1]) cntL[i] = a[i - 1] - mn;
        if(a[i + n - 1] + cntL[i] >= a[i - 1]) cntR[i] = a[i + n - 1] + cntL[i] - a[i - 1];
        if(cntL[i] + cntR[i] < need) need = cntL[i] + cntR[i];
    }
    int who = -1;
    for(int i = 1; i <= n; i++) {
        if(cntL[i] + cntR[i] > need) continue;
        if(who == -1) who = i;
        else {
            if(check(i, who, need + n)) who = i;
        }
    }
    while(cntL[who]--) putchar(‘(‘);
    for(int i = who; i < who + n; i++) putchar(s[i]);
    while(cntR[who]--) putchar(‘)‘);
    puts("");
    return 0;
}

/**/

原文地址：https://www.cnblogs.com/CJLHY/p/10990538.html