QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#532140#2343. First of Her NameSwarthmore#WA 0ms3796kbC++202.7kb2024-08-25 00:33:052024-08-25 00:33:05

Judging History

你现在查看的是最新测评结果

  • [2024-08-25 00:33:05]
  • 评测
  • 测评结果:WA
  • 用时:0ms
  • 内存:3796kb
  • [2024-08-25 00:33:05]
  • 提交

answer

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;

#define FOR(i, a, b) for (int i = a; i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0Rd(i, a) for (int i = (a) - 1; i >= 0; i--)
#define trav(a, x) for (auto &a : x)
#define sz(x) (int)(x).size()
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert

const char nl = '\n';

const int MX = 1e6+1;
const int L = 21;
int N;
int Q;
char C[MX];
int P[MX];

int get(string &S, int v) {
    F0R(i, sz(S)) {
        if (v == -1) return 1;
        if (S[i] != C[v]) {
            if (S[i] < C[v]) return -1;
            return 1;
        }
        v = P[v];
    }
    return 0;
}

void solve() {
    cin >> N >> Q;
    F0R(i, N) cin >> C[i] >> P[i];
    F0R(i, N) P[i]--;
    int anc[N][22]; F0R(i, N) {
        anc[i][0] = P[i];
    }
    FOR(j, 1, 22) {
        F0R(i, N) {
            if (anc[i][j-1] == -1) {
                anc[i][j] = -1;
            } else {
                anc[i][j] = anc[anc[i][j-1]][j-1];
            }
        }
    }

    int cla[N]; F0R(i, N) cla[i] = C[i] - 'A';
    F0R(iter, 22) {
        vector<array<int, 3>> cur;
        F0R(i, N) {
            cur.pb({cla[i], (anc[i][iter]==-1?-1:cla[anc[i][iter]]), i});
        }
        sort(all(cur));
        int lst = 0;
        F0R(i, N) {
            if (i && (cur[i][0] != cur[i-1][0] || cur[i][1] != cur[i-1][1])) {
                lst = i;
            }
            cla[cur[i][2]] = lst;
        }
        /*F0R(i, N) {
            if (iter == 0) {
                cout << i << ": " << cla[i] << nl;
            }
        }*/

    }

    vpi ord;
    F0R(i, N) {
        ord.pb({cla[i], i});
    }
    sort(all(ord));
    /*F0R(i, N) {
        cout << ord[i].s << nl;
    }
    cout << nl;*/

    while(Q--) {
        string S; cin >> S;
        // find first geq S 
        int lo1 = 0, hi1 = N-1;
        while (lo1 < hi1) {
            int mid = (lo1+hi1)/2;
            if (get(S, ord[mid].s) <= 0) {
                hi1 = mid;
            } else lo1 = mid+1;
        }
        int lo2 = 0, hi2 = N-1;
        // find last leq S
        while (lo2 < hi2) {
            int mid = (lo2+hi2+1)/2;
            if (get(S, ord[mid].s) >= 0) {
                lo2 = mid;
            } else hi2 = mid-1;
        }
        cout << lo2-lo1+1 << nl;
    }
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
	solve();
	return 0;
}

Details

Test #1:

score: 100
Accepted
time: 0ms
memory: 3624kb

Test #2:

score: 0
Accepted
time: 0ms
memory: 3796kb

Test #3:

score: 0
Accepted
time: 0ms
memory: 3628kb

Test #4:

score: -100
Wrong Answer
time: 0ms
memory: 3564kb