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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#707716#8642. Spy 3hhoppitree0 8ms5756kbC++178.3kb2024-11-03 17:11:492024-11-03 17:11:51

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你现在查看的是最新测评结果

  • [2024-11-03 17:11:51]
  • 评测
  • 测评结果:0
  • 用时:8ms
  • 内存:5756kb
  • [2024-11-03 17:11:49]
  • 提交

Aoi

#include <bits/stdc++.h>
#include "Aoi.h"

using namespace std;

vector<int> operator * (vector<int> x, int y) {
    if (x.empty()) return {};
    vector<int> tres(x.size());
    for (int i = 0; i < (int)x.size(); ++i) {
        tres[i] += x[i] * y;
        if (i + 1 != (int)x.size()) {
            tres[i + 1] += tres[i] >> 10;
            tres[i] &= 1023;
        }
    }
    while (tres.back() >= 1024) {
        tres.push_back(tres.back() >> 10);
        tres[tres.size() - 2] &= 1023;
    }
    return tres;
}

vector<int> operator + (vector<int> x, int y) {
    if (!y) return x;
    if (x.empty()) {
        x = {0};
    }
    x[0] += y;
    int r = 0;
    while (r < (int)x.size() && x[r] >= 1024) {
        x[r] -= 1024;
        if (r == (int)x.size() - 1) x.push_back(0);
        ++x[r + 1];
    }
    return x;
}

string tr(vector<int> arr) {
    string res;
    for (int i = (int)arr.size() - 1; ~i; --i) {
        for (int j = 9; ~j; --j) {
            res += (char)((arr[i] >> j) + '0');
        }
    }
    while (!res.empty() && res[0] == '0') {
        res = res.substr(1);
    }
    return res;
}

namespace Aoi {
    const int N = 1e4 + 5;

    vector< pair<int, long long> > G[N];
    long long dis[N];
    int pre[N];
    vector<int> T[N];
    int dep[N], clk[N];

    void dfs(int x) {
        for (auto v : T[x]) {
            dep[v] = dep[x] + 1, dfs(v), clk[x] = min(clk[x], clk[v]);
        }
    }

    int LCA(int x, int y) {
        while (x != y) {
            if (dep[x] < dep[y]) swap(x, y);
            x = pre[x];
        }
        return x;
    }

    map< pair<int, int>, int> spe;

    int getId(int x) {
        while (x) {
            if (spe.count({x, pre[x]})) {
                return spe[{x, pre[x]}] + 1;
            }
            x = pre[x];
        }
        return 0;
    }

    string calc(vector< pair<int, int> > o) {
        vector<int> res;
        for (int i = (int)o.size() - 1; ~i; --i) {
            res = res * o[i].first + o[i].second;
        }
        return tr(res);
    }

    string aoi(int n, int m, int q, int k, vector<int> ox, vector<int> oy, vector<long long> oz, vector<int> qr, vector<int> o) {
        for (int i = 0; i < m; ++i) {
            G[ox[i]].push_back({oy[i], oz[i]});
            G[oy[i]].push_back({ox[i], oz[i]});
        }
        for (int i = 1; i < n; ++i) {
            dis[i] = 1e18;
        }
        priority_queue< pair<long long, int> > pq;
        pq.push({0, 0});
        while (!pq.empty()) {
            long long D = -pq.top().first;
            int x = pq.top().second;
            pq.pop();
            if (D != dis[x]) continue;
            for (auto [v, w] : G[x]) {
                long long nD = D + w;
                if (nD < dis[v] || (nD == dis[v] && x < pre[v])) {
                    dis[v] = nD, pre[v] = x;
                    if (nD != dis[v]) pq.push({-nD, v});
                }
            }
        }
        for (int i = 0; i < n; ++i) clk[i] = q;
        for (int i = 0; i < q; ++i) {
            clk[qr[i]] = min(clk[qr[i]], i);
        }
        for (int i = 1; i < n; ++i) {
            T[pre[i]].push_back(i);
        }
        dfs(0);
        vector< pair<int, int> > res;
        for (int i = 0; i < k; ++i) {
            if (dep[ox[o[i]]] < dep[oy[o[i]]]) swap(ox[o[i]], oy[o[i]]);
            res.push_back({q + 1, clk[ox[o[i]]]});
            spe[{ox[o[i]], oy[o[i]]}] = spe[{ox[o[i]], oy[o[i]]}] = i;
        }
        for (int i = 1; i < q; ++i) {
            res.push_back({k + 1, getId(LCA(qr[i - 1], qr[i]))});
        }
        string ret = calc(res);
        assert((int)ret.size() <= 1350);
        return ret;
    }
}

string aoi(int n, int m, int q, int k, vector<int> ox, vector<int> oy, vector<long long> oz, vector<int> qr, vector<int> o) {
    return Aoi::aoi(n, m, q, k, ox, oy, oz, qr, o);
}

Bitaro

#include <bits/stdc++.h>
#include "Bitaro.h"

using namespace std;

vector<int> operator / (vector<int> x, int y) {
    vector<int> tres(x.size());
    int now = 0;
    for (int i = (int)x.size() - 1; ~i; --i) { 
        now = (now << 10) + x[i];
        tres[i] = now / y;
        now %= y;
    }
    while (!tres.empty() && !tres.back()) {
        tres.pop_back();
    }
    return tres;
}

int operator % (vector<int> x, int y) {
    vector<int> tres(x.size());
    int now = 0;
    for (int i = (int)x.size() - 1; ~i; --i) { 
        now = (now << 10) + x[i];
        tres[i] = now / y;
        now %= y;
    }
    return now;
}

vector<int> parse(string S) {
    while (S.size() % 10) S = "0" + S;
    vector<int> res(S.size() / 10);
    for (int i = 0; i < (int)S.size(); i += 10) {
        for (int j = 0; j < 10; ++j) {
            res[(int)S.size() - 1 - i / 10] += (S[i + j] - '0') << (9 - j);
        }
    }
    return res;
}

namespace Bitaro {
    const int N = 1e4 + 5;

    vector<int> apd[N];
    int pre[N], det[N], dep[N], usd[N];
    vector< tuple<int, int, long long> > E;
    long long dis[N];
    vector< pair<int, int> > ts;
    vector< pair<int, long long> > G[N];

    vector<int> process(int n, vector<int> imp, int id) {
        for (int i = 0; i < n; ++i) {
            G[i].clear();
            dis[i] = (i == 0 ? 0 : 1e18), pre[i] = 0;
        }
        for (auto [x, y, z] : E) {
            G[x].push_back({y, z});
            G[y].push_back({x, z});
        }
        for (auto x : imp) {
            G[ts[x].first].push_back({ts[x].second, 0});
            G[ts[x].second].push_back({ts[x].first, 0});
        }
        priority_queue< pair<long long, int> > pq;
        pq.push({0, 0});
        while (!pq.empty()) {
            long long D = -pq.top().first;
            int x = pq.top().second;
            pq.pop();
            if (D != dis[x]) continue;
            for (auto [v, w] : G[x]) {
                long long nD = D + w;
                if (nD < dis[v] || (nD == dis[v] && x < pre[v])) {
                    dis[v] = nD, pre[v] = x;
                    if (nD != dis[v]) pq.push({-nD, v});
                }
            }
        }
        vector<int> path = {id};
        int now = id;
        while (now) {
            path.push_back(pre[now]);
            now = pre[now];
        }
        reverse(path.begin(), path.end());
        for (int i = 1; i < (int)path.size(); ++i) {
            det[path[i]] = path[i - 1];
            dep[path[i]] = dep[path[i - 1]] + 1;
        }
        return path;
    }

    void bitaro(int n, int m, int q, int k, vector<int> ox, vector<int> oy, vector<long long> oz, vector<int> qr, vector<int> o, string S) {
        vector<int> val = parse(S);
        for (int i = 0; i < k; ++i) {
            int cl = val % (q + 1);
            val = val / (q + 1);
            apd[cl].push_back(i);
        }
        map< pair<int, int>, int> M;
        for (int i = 0; i < m; ++i) {
            M[{ox[i], oy[i]}] = M[{oy[i], ox[i]}] = i;
        }
        for (int i = 0; i < k; ++i) {
            usd[o[i]] = 1;
            ts.push_back({ox[o[i]], oy[o[i]]});
        }
        for (int i = 0; i < m; ++i) {
            if (!usd[i]) E.push_back({ox[i], oy[i], oz[i]});
        }
        for (int i = 0; i < q; ++i) {
            vector<int> imp;
            if (i) {
                int id = val % (k + 1);
                val = val / (k + 1);
                if (dep[ox[o[id]]] < dep[oy[o[id]]]) {
                    swap(dep[ox[o[id]]], dep[oy[o[id]]]);
                }
                id = ox[o[id]];
                while (id) {
                    if (usd[M[{id, pre[i]}]]) {
                        imp.push_back(M[{id, pre[id]}]);
                    }
                    id = det[id];
                }
            }
            for (auto x : apd[i]) {
                imp.push_back(x);
            }
            vector<int> res = process(n, imp, qr[i]);
            vector<int> ids;
            for (int j = 0; j + 1 < (int)res.size(); ++j) {
                ids.push_back(M[{res[j], res[j + 1]}]);
            }
            answer(ids);
        }
    }
}

void bitaro(int n, int m, int q, int k, vector<int> ox, vector<int> oy, vector<long long> oz, vector<int> qr, vector<int> o, string S) {
    Bitaro::bitaro(n, m, q, k, ox, oy, oz, qr, o, S);
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 0
Wrong Answer
time: 8ms
memory: 5756kb

Manager to Aoi

200 19900
13 70 985302938314
120 174 18964037101
18 153 170196070829
45 129 323777973024
62 198 689223413645
88 133 457404464825
19 57 803409835578
22 187 662331177910
18 31 529437059733
161 182 637731822589
109 131 32831735773
109 191 875742441191
43 78 135479410688
56 60 19000632823
44 143 6823771...

Aoi to Manager

1

Manager to Bitaro

WA

Bitaro to Manager

-1

Manager to Checker

0.00

result:

points 0.0