QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#642105#4811. Be CarefulhhoppitreeRE 0ms0kbC++174.0kb2024-10-15 10:21:342024-10-15 10:21:35

Judging History

你现在查看的是最新测评结果

  • [2024-10-15 10:21:35]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2024-10-15 10:21:34]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

const int N = 205, P = 998244353;

struct MOD {
    int operator ()(long long x) {
        return x - (((__int128)x * 18479187002) >> 64) * P;
    }
} mod;

vector<int> G[N];
int n, dp[N][N], sdp[N][N], f[2][1 << 18], g[2][1 << 18][N], C[N][N];

void dfs(int x, int fa = -1) {
    int deg = 0, cnt = 0;
    for (auto v : G[x]) {
        if (v != fa) {
            dfs(v, x);
            for (int i = n; i >= 0; --i) {
                sdp[v][i] = mod(sdp[v][i + 1] + dp[v][i]);
            }
            cnt += (G[v].size() == 1), ++deg;
        }
    }
    if (!deg) {
        fill(dp[x], dp[x] + n + 1, 1);
        return;
    }
    vector< pair<int, int> > o;
    for (auto v : G[x]) {
        if (v != fa && G[v].size() != 1) {
            int L = n;
            while (!dp[v][L]) --L;
            o.push_back({L, v});
        }
    }
    sort(o.begin(), o.end());
    int mn = o.size() + 1, ps = -1;
    for (int i = 0; i < (int)o.size(); ++i) {
        if (o[i].first + ((int)o.size() - i) < mn) {
            mn = o[i].first + ((int)o.size() - i), ps = i;
        }
    }
    int tz = (!~ps ? 0 : o[ps].first + 1);
    for (int i = 0; i < 1 << tz; ++i) f[0][i] = !i;
    for (int i = 0; i <= ps; ++i) {
        for (int j = 0; j < 1 << tz; ++j) f[1][j] = f[0][j], f[0][j] = 0;
        for (int j = (1 << (o[i].first + 1)) - 1; ~j; --j) {
            for (int k = 0; k <= o[i].first; ++k) {
                f[0][j | (1 << k)] = mod(f[0][j | (1 << k)] + 1ll * f[1][j] * dp[o[i].second][k]);
            }
        }
    }
    for (int i = 0; i < 1 << tz; ++i) if (f[0][i]) {
        int z = o.size() - 1 - ps;
        for (int j = 0; j < (1 << z); ++j) {
            for (int k = 0; k <= cnt; ++k) g[0][j][k] = 0;
        }
        g[0][(1 << z) - 1][cnt] = f[0][i];
        for (int j = 0; ; ++j) {
            int ok = 0;
            for (int k = 0; k < (1 << z); ++k) {
                for (int l = 0; l <= cnt; ++l) {
                    if (g[0][k][l]) ok = 1;
                    g[1][k][l] = g[0][k][l], g[0][k][l] = 0;
                }
            }
            if (!ok) break;
            for (int l = 0; l <= cnt; ++l) {
                for (int k = 0; k < (1 << z); ++k) if (g[1][k][l]) {
                    if (j >= 30 || !((i >> j) & 1)) {
                        ((g[0][k][l] -= g[1][k][l]) += P) && (g[0][k][l] += P);
                        int mul = g[1][k][l];
                        for (int t = 1; t <= l; ++t) mul = mod(1ll * mul * (n - j));
                        for (int p = 0; p < z; ++p) {
                            if ((k >> p) & 1) {
                                mul = mod(1ll * mul * sdp[o[ps + 1 + p].second][j + 1]);
                            }
                        }
                        dp[x][j] = mod(dp[x][j] + mul);
                    }
                }
                for (int len = 2; len <= (1 << z); len <<= 1) {
                    int pos = o[ps + __builtin_ctz(len)].second;
                    for (int k = 0; k < (1 << z); k += len) {
                        for (int p = k; p < k + (len >> 1); ++p) {
                            g[1][p][l] = mod(g[1][p][l] + 1ll * g[1][p + (len >> 1)][l] * dp[pos][j]);
                        }
                    }
                }
                for (int k = 0; k < (1 << z); ++k) {
                    for (int L = 0; L <= l; ++L) {
                        g[0][k][L] = mod(g[0][k][L] + 1ll * g[1][k][l] * C[l][L]);
                    }
                }
            }
        }
    }
}

signed main() {
    scanf("%d", &n);
    for (int i = C[0][0] = 1; i <= n; ++i) {
        for (int j = C[i][0] = 1; j <= i; ++j) {
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
        }
    }
    for (int i = 1, x, y; i < n; ++i) {
        scanf("%d%d", &x, &y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    dfs(1);
    for (int i = 0; i <= n; ++i) printf("%d\n", (dp[1][i] % P + P) % P);
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 0
Runtime Error

input:

5
1 2
1 3
2 4
2 5

output:


result: