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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#134744#2443. Dense Subgraphckiseki#AC ✓330ms21228kbC++233.6kb2023-08-04 19:24:592023-08-04 19:25:03

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-08-04 19:25:03]
  • 评测
  • 测评结果:AC
  • 用时:330ms
  • 内存:21228kb
  • [2023-08-04 19:24:59]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

#define all(x) begin(x), end(x)
#ifdef CKISEKI
#define safe cerr << __PRETTY_FUNCTION__ << " line " << __LINE__ << " safe\n"
#define debug(a...) debug_(#a, a)
#define orange(a...) orange_(#a, a)
template <typename ...T>
void debug_(const char *s, T ...a) {
    cerr << "\e[1;32m(" << s << ") = (";
    int cnt = sizeof...(T);
    (..., (cerr << a << (--cnt ? ", " : ")\e[0m\n")));
}
template <typename I>
void orange_(const char *s, I L, I R) {
    cerr << "\e[1;32m[ " << s << " ] = [ ";
    for (int f = 0; L != R; L++)
        cerr << (f++ ? ", " : "") << *L;
    cerr << " ]\e[0m\n";
}
#else 
#define safe ((void)0)
#define debug(...) safe
#define orange(...) safe
#endif

namespace {

constexpr int kN = 35000 + 5;
constexpr int kMod = 1'000'000'007;
constexpr int kD = 1 << 5;

constexpr int add(int a, int b) {
    return a + b >= kMod ? a + b - kMod : a + b;
}
constexpr int mul(int64_t a, int64_t b) {
    return static_cast<int>(a * b % kMod);
}

int a[kN];
vector<int> g[kN];

void pre(int u, int f) {
    vector<int> cur;
    for (int v : g[u]) {
        if (v == f)
            continue;
        cur.push_back(v);
        pre(v, u);
    }
    g[u] = cur;
}

int sm[kN][kD];
int dp[kN][kD + 1];
bool bad[kN][kD];

int dfs(int u) {
    const int d = int(g[u].size());
    const int d2 = 1 << d;
    sm[u][0] = a[u];
    for (int S = 1; S < d2; ++S) {
        sm[u][S] = a[u];
        for (int i = 0; i < d; ++i) {
            if ((S >> i) & 1) {
                sm[u][S] += a[g[u][i]];
                bad[u][S] |= bad[u][S ^ (1 << i)];
            }
        }
        if (sm[u][S] > 0)
            bad[u][S] = true;
    }

    dp[u][kD] = 1;
    for (int v : g[u]) {
        dp[u][kD] = mul(dp[u][kD], dfs(v));
    }
    vector<int> tot(g[u].size());
    for (int i = 0; i < d; ++i) {
        const int v = g[u][i];
        const int vd2 = 1 << int(g[v].size());
        vector<bool> badbad(vd2);
        for (int j = 0; j < vd2; ++j) {
            badbad[j] = badbad[j] | bad[v][j];
            badbad[j] = badbad[j] | bool(sm[v][j] + a[u] > 0);
            for (int k = 0; k < int(g[v].size()); ++k)
                if ((j >> k) & 1)
                    badbad[j] = badbad[j] | bool(badbad[j ^ (1 << k)]);
            if (badbad[j])
                continue;
            tot[i] = add(tot[i], dp[v][j]);
        }
    }
    int ret = dp[u][kD];
    for (int S = 0; S < d2; ++S) {
        if (bad[u][S])
            continue;
        dp[u][S] = 1;
        for (int j = 0; j < d; ++j)
            if ((S >> j) & 1)
                dp[u][S] = mul(dp[u][S], tot[j]);
            else
                dp[u][S] = mul(dp[u][S], dp[g[u][j]][kD]);
        ret = add(ret, dp[u][S]);
    }
    return ret;
}

} // namespace

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        int n, x;
        cin >> n >> x;
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
            a[i] -= x;
        }
        for (int i = 1; i < n; ++i) {
            int u, v;
            cin >> u >> v;
            u -= 1, v -= 1;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        pre(0, 0);
        cout << dfs(0) << '\n';
        for (int i = 0; i < n; ++i) {
            memset(dp[i], 0, sizeof(dp[i]));
            memset(sm[i], 0, sizeof(sm[i]));
            memset(bad[i], 0, sizeof(bad[i]));
            g[i].clear();
        }
    }
    return 0;
}
/*
1
5 0
1 1 1 1 1
1 2
2 3
3 4
4 5
bad[0][1] = true (2)
bad[1][1] = true (2)
bad[2][1] = true (2)
bad[3][1] = true (2)
dp[4][No] = 1
dp[3][No] = 2
dp[3][1] = 0
dp[2][No] = 3
dp[2][1] = 0
dp[1][No] = 4
dp[1][1] = 0
dp[0][No] = 5
dp[0][1] = 0
6
*/

详细

Test #1:

score: 100
Accepted
time: 330ms
memory: 21228kb

input:

30
10 11086
10189 24947 2265 9138 27104 12453 15173 3048 30054 2382
8 1
1 4
5 10
10 4
3 5
2 10
9 7
6 10
7 1
15 9664
4127 24649 13571 8586 34629 8644 3157 33133 3713 32646 29412 8108 13583 21362 23735
14 9
7 1
15 12
10 15
2 6
3 11
9 1
1 11
6 12
4 10
13 15
8 15
12 11
5 3
20 29310
21738 9421 8412 4617 ...

output:

320
3312
1048576
60461799
663660496
831386303
528321417
945912389
820338255
58962594
643816787
354510769
532631871
280169661
533022884
475656636
892230988
381315031
40006857
376652471
288541404
546717513
976236489
615823216
419541488
840899550
546169620
634600456
825179565
169850560

result:

ok 30 lines