QOJ.ac

QOJ

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
#106619#6402. MEXimum Spanning Treetom0727WA 1ms4348kbC++146.0kb2023-05-18 12:15:052023-05-18 12:15:52

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.
  • [2023-05-18 12:15:52]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:4348kb
  • [2023-05-18 12:15:05]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#if __cplusplus >= 201103L
    struct pairhash {
        static uint64_t splitmix64(uint64_t x) {
            x += 0x9e3779b97f4a7c15;
            x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
            x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
            return x ^ (x >> 31);
        }

        template<class T, class U>
        size_t operator() (const pair<T,U> &p) const {
            static const uint64_t FIXED_RANDOM = (uint64_t)chrono::steady_clock::now().time_since_epoch().count();
            return splitmix64(p.first + FIXED_RANDOM) ^ splitmix64(p.second+ FIXED_RANDOM);
        }
    };
    struct custom_hash {
        static uint64_t splitmix64(uint64_t x) {
            x += 0x9e3779b97f4a7c15;
            x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
            x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
            return x ^ (x >> 31);
        }

        size_t operator()(uint64_t x) const {
            static const uint64_t FIXED_RANDOM = (uint64_t)chrono::steady_clock::now().time_since_epoch().count();
            return splitmix64(x + FIXED_RANDOM);
        }
    };
    mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
    inline int randint(int l, int r) {
        return uniform_int_distribution<int>(l,r)(rng);
    }
    inline double randdouble(double l, double r) {
        return uniform_real_distribution<double>(l,r)(rng);
    }
#endif
 
#ifndef ONLINE_JUDGE
#  define LOG(x) (cerr << #x << " = " << (x) << endl)
#else
#  define LOG(x) 0
#endif

#define fastio ios::sync_with_stdio(false); cin.tie(0);
#define ll long long
#define ull unsigned long long
#define ll128 __int128_t
#define PI 3.14159265358979323846
#define abs(a) ((a>0)?a:-(a))
#define pii pair<int,int>
#define pll pair<ll,ll>

const long double pi = acos(-1.0);
const long double eps = (double)1e-2;

int mod = (1<<30);

template<class T>
T qpow(T a, int b) {
    T res = 1;
    while (b) {
        if (b & 1) res *= a;
        a *= a;
        b >>= 1;
    }
    return res;
}
int norm(int x) {
    if (x < 0) {
        x += mod;
    }
    if (x >= mod) {
        x -= mod;
    }
    return x;
}
struct Z {
    int x;
    Z(int x = 0) : x(norm(x)) {}
    Z(ll x) : x(norm((int)(x % mod))) {}
    int val() const {
        return x;
    }
    Z operator-() const {
        return Z(norm(mod - x));
    }
    Z inv() const {
        assert(x != 0);
        return qpow(*this, mod - 2);
    }
    Z &operator*=(const Z &rhs) {
        x = (ll)(x) * rhs.x % mod;
        return *this;
    }
    Z &operator+=(const Z &rhs) {
        x = norm(x + rhs.x);
        return *this;
    }
    Z &operator-=(const Z &rhs) {
        x = norm(x - rhs.x);
        return *this;
    }
    Z &operator/=(const Z &rhs) {
        return *this *= rhs.inv();
    }
    friend Z operator*(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res *= rhs;
        return res;
    }
    friend Z operator+(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res += rhs;
        return res;
    }
    friend Z operator-(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res -= rhs;
        return res;
    }
    friend Z operator/(const Z &lhs, const Z &rhs) {
        Z res = lhs;
        res /= rhs;
        return res;
    }
    friend std::istream &operator>>(std::istream &is, Z &a) {
        ll v;
        is >> v;
        a = Z(v);
        return is;
    }
    friend std::ostream &operator<<(std::ostream &os, const Z &a) {
        return os << a.val();
    }
};

const int maxn = 1e4+5;
const int maxm = 4e4+55;

struct State {
    int u, v, szu, szv;
} st[maxm];  // 注意这里是 maxm,应该是边的数量
int uni = 0;  // union 的次数,如果等于 n-1 说明联通了
int n, m, k;
struct DSU {
    int par[maxn], sz[maxn], tail = 0;
    inline void init() {
        for (int i = 1; i <= n; i++) par[i] = i, sz[i] = 1;
        tail = 0;
    }
    int finds(int u) {
        if (par[u] == u) return u;
        return finds(par[u]);  // 无路径压缩
    }
    void unions(int u, int v) {
        u = finds(u), v = finds(v);
        if (sz[u] < sz[v]) swap(u,v);  // sz[u] >= sz[v]
        st[++tail] = {u, v, sz[u], sz[v]};   // 无论是否 union 成功都要push到 stack 里
        if (u == v) return;
        par[v] = u;
        sz[u] += sz[v];
        uni++;  // 成功union
    }
    void cancel() {
        if (tail > 0) {
            int u = st[tail].u, v = st[tail].v;
            par[v] = v;
            if (sz[u] != st[tail].szu) uni--;  // 成功回退一次union
            sz[u] = st[tail].szu;
            sz[v] = st[tail].szv;
            tail--;
        }
    }
} dsu;

struct Edge {
    int u, v, w;
};
struct TreeNode {
    vector<Edge> vec;
};

int ans = 0;
struct SegmentTree {
    TreeNode tr[maxn<<2];
    void insert(int cur, int l, int r, int L, int R, Edge e) {
        if (L <= l && R >= r) {
            tr[cur].vec.push_back(e);
            return;
        }
        int mid = (l+r) >> 1;
        if (L <= mid) insert(cur<<1, l, mid, L, R, e);
        if (R > mid) insert(cur<<1|1, mid+1, r, L, R, e);
    }
    void dfs(int cur, int l, int r) {
        for (Edge& e : tr[cur].vec) {
            dsu.unions(e.u, e.v);
            dsu.unions(e.u, e.v);
        }

        if (l == r) {
            if (uni == n-1) ans = max(ans, l);
        } else {
            int mid = (l+r) >> 1;
            dfs(cur<<1, l, mid);
            dfs(cur<<1|1, mid+1, r);
        }

        for (Edge& e : tr[cur].vec) {
            dsu.cancel();
            dsu.cancel();
        }
    }
} tr;

int main() {
    fastio;
    cin >> n >> m;
    dsu.init();
    int N = 1e3;
    for (int i = 1; i <= m; i++) {
        int u, v, w; cin >> u >> v >> w;
        if (w > 0)
            tr.insert(1, 0, N, 0, w-1, {u, v, w});
        if (w < N)
            tr.insert(1, 0, N, w+1, N, {u, v, w});
    }
    tr.dfs(1, 0, N);
    cout << ans << "\n";
}

詳細信息

Test #1:

score: 0
Wrong Answer
time: 1ms
memory: 4348kb

input:

4 4
1 2 0
2 3 1
1 3 1
3 4 2

output:

1000

result:

wrong answer 1st numbers differ - expected: '3', found: '1000'