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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
#698133#9431. The Quest for El DoradoHojstyerRE 0ms3656kbC++204.9kb2024-11-01 17:35:502024-11-01 17:35:55

Judging History

你现在查看的是最新测评结果

  • [2024-11-01 17:35:55]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:3656kb
  • [2024-11-01 17:35:50]
  • 提交

answer

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

typedef long long ll;

template <typename T>
struct ST
{
    T MAX = 1e9;
    vector<vector<T>> maxn;
    vector<T> a;
    const int m = 20;
    vector<T> LOG;

    ST() {}
    ST(int n, vector<T> b)
    {
        a.resize(n + 1, 0);
        a = b;
        init(n);
    }

    void init(int n)
    {
        if (std::is_same<T, ll>::value)
            MAX = 1e18;
        maxn.resize(n + 1, vector<T>(m + 1, -MAX));
        LOG.resize(n + 1, 0);
        for (int i = 1; i <= n; i++)
            maxn[i][0] = a[i];
        LOG[1] = 0;
        for (int i = 2; i <= n; i++)
            LOG[i] = LOG[i / 2] + 1;
        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 1; i + (1 << (j - 1)) <= n; i++)
                maxn[i][j] = max(maxn[i][j - 1], maxn[i + (1 << (j - 1))][j - 1]);
    }

    void init(int n, vector<T> b)
    {
        if (std::is_same<T, ll>::value)
            MAX = 1e18;
        a.resize(n + 1, 0);
        a = b;
        maxn.resize(n + 1, vector<T>(m + 1, -MAX));
        LOG.resize(n + 1, 0);
        for (int i = 1; i <= n; i++)
            maxn[i][0] = a[i];
        LOG[1] = 0;
        for (int i = 2; i <= n; i++)
            LOG[i] = LOG[i / 2] + 1;
        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 1; i + (1 << (j - 1)) <= n; i++)
                maxn[i][j] = max(maxn[i][j - 1], maxn[i + (1 << (j - 1))][j - 1]);
    }

    T query_max(int l, int r)
    {
        int d = LOG[r - l + 1];
        return max(maxn[l][d], maxn[r - (1 << d) + 1][d]);
    }
};

void solve()
{
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<array<int, 3>>> adj(n + 1);
    for (int i = 1; i <= m; ++i)
    {
        int u, v, c, l;
        cin >> u >> v >> c >> l;
        adj[u].push_back({v, c, l});
        adj[v].push_back({u, c, l});
    }

    vector<int> a(n + 1), b(n + 1);
    vector<vector<int>> pos(m + 1);
    for (int i = 1; i <= k; ++i)
    {
        cin >> a[i] >> b[i];
        pos[a[i]].push_back(i);
    }

    vector<ST<int>> seg(m + 1);
    for (int i = 1; i <= m; ++i)
    {
        int r = pos[i].size();
        vector<int> f(r + 1);
        for (int j = 0; j < r; ++j)
            f[j + 1] = b[pos[i][j]];
        seg[i].init(r + 1, f);
    }

    priority_queue<array<ll, 3>> q;
    vector<ll> dis(n + 1, 1e18), tg(n + 1, k + 1);
    dis[1] = 0, tg[1] = 0;
    q.push({0, -dis[1], 1});
    while (!q.empty())
    {
        auto [nk, _, u] = q.top();
        q.pop();
        for (auto [v, p, w] : adj[u])
        {
            if (a[tg[u]] == p)
            {
                ll np = tg[u];
                // tg[v] = tg[u], dis[v] = dis[u] + w;
                if (dis[u] + w > b[np])
                {
                    int j = lower_bound(pos[a[np]].begin(), pos[a[np]].end(), np + 1) - pos[a[np]].begin();
                    int l = j + 1, r = pos[a[np]].size(), now = -1;
                    while (l <= r)
                    {
                        int mid = l + r >> 1;
                        if (seg[a[np]].query_max(l, mid) >= w)
                            r = mid - 1, now = mid;
                        else
                            l = mid + 1;
                    }
                    if (now != -1 && (pos[a[np]][now - 1] < tg[v] || (pos[a[np]][now - 1] == tg[v] && dis[v] > w)))
                    {
                        tg[v] = pos[a[np]][now - 1];
                        dis[v] = w;
                        q.push({-tg[v], -dis[v], v});
                    }
                }
                else if (tg[u] < tg[v] || (tg[u] == tg[v] && dis[v] > dis[u] + w))
                {
                    tg[v] = tg[u], dis[v] = dis[u] + w;
                    q.push({-tg[v], -dis[v], v});
                }
            }
            else
            {
                int j = lower_bound(pos[p].begin(), pos[p].end(), tg[u] + 1) - pos[p].begin();
                int l = j + 1, r = pos[p].size(), now = -1;
                while (l <= r)
                {
                    int mid = l + r >> 1;
                    if (seg[p].query_max(l, mid) >= w)
                        r = mid - 1, now = mid;
                    else
                        l = mid + 1;
                }
                if (now != -1 && (pos[p][now - 1] < tg[v] || (pos[p][now - 1] == tg[v] && dis[v] > w)))
                {
                    tg[v] = pos[p][now - 1];
                    dis[v] = w;
                    q.push({-tg[v], -dis[v], v});
                }
            }
        }
    }

    string ans;
    for (int i = 1; i <= n; ++i)
        ans += (tg[i] > k ? '0' : '1');
    cout << ans << "\n";
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3656kb

input:

2
5 6 4
1 2 1 30
2 3 1 50
2 5 5 50
3 4 6 10
2 4 5 30
2 5 1 40
1 70
6 100
5 40
1 30
3 1 1
2 3 1 10
1 100

output:

11011
100

result:

ok 2 lines

Test #2:

score: -100
Runtime Error

input:

1110
46 80 30
44 23 5 46
10 28 1 64
32 34 3 40
9 36 1 26
15 14 5 95
38 19 2 34
2 17 4 183
10 38 2 81
5 15 2 83
31 38 3 100
40 30 1 53
41 10 1 193
29 20 5 18
14 41 3 78
8 16 5 74
46 13 3 78
44 28 3 45
1 40 3 133
5 32 1 108
22 26 2 83
10 38 1 77
11 40 1 11
17 21 2 66
41 46 3 98
9 36 2 25
40 18 1 19
27...

output:


result: