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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
#572743#9319. Bull FarmKeeperHihiTL 93ms24700kbC++203.5kb2024-09-18 16:12:192024-09-18 16:12:19

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 16:12:19]
  • 评测
  • 测评结果:TL
  • 用时:93ms
  • 内存:24700kb
  • [2024-09-18 16:12:19]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

struct Info {
    int cur = 0;
    int mx = -1e9;
    friend bool operator < (Info a, Info b) {
        return a.mx > b.mx;
    }
};

int cal(char x, char y) {
    return (x - 48) * 50 + y - 48;
}

void solve() {
    int n, l, q;
    cin >> n >> l >> q;
    vector to(l + 1, vector<int>(n + 1));
    vector inv(l + 1, vector<int>(n + 1));
    vector<vector<pair<int, int>>> adj(n + 1);
    for (int i = 1; i <= l; i++) {
        string s;
        cin >> s;
        vector<int> vis(n + 1);
        for (int j = 0, k = 1; j < s.size(); j += 2, k++) {
            to[i][k] = cal(s[j], s[j + 1]);
            inv[i][to[i][k]] = k;
            vis[to[i][k]] = 1;
        }
        int cnt = accumulate(vis.begin() + 1, vis.end(), 0);
        if (cnt == n) {
            for (int j = 1; j <= n; j++) {
                adj[j].emplace_back(to[i][j], i);
                adj[to[i][j]].emplace_back(j, i);
            }
        } else if (cnt == n - 1) {
            int x = 0;
            vis.assign(n + 1, 0);
            for (int j = 1; j <= n; j++) {
                if (vis[to[i][j]]) {
                    x = to[i][j];
                    break;
                }
                vis[to[i][j]] = 1;
            }
            vector<int> a;
            for (int j = 1; j <= n; j++) {
                if (to[i][j] == x) {
                    a.emplace_back(j);
                }
            }
            assert(a.size() == 2);
            // 初始只能在 a[0] 或者 a[1],其结果都是唯一确定的,把对应的几个点 merge 一下就行了

            auto dfs = [&](auto self, int u, int fa, int t) -> void {
                if (vis[u]) {
                    // cout << "edge = " << t << ' ' << fa << endl;
                    adj[t].emplace_back(fa, i);
                    return;
                }
                vis[u] = 1;
                self(self, inv[i][u], u, t);
            };
            vis.assign(n + 1, 0);
            vis[0] = 1;
            inv[i][x] = a[1];
            dfs(dfs, a[0], a[0], a[0]);
            vis.assign(n + 1, 0);
            vis[0] = 1;
            inv[i][x] = a[0];
            dfs(dfs, a[1], a[1], a[1]);
            // 注意是单向边!!!
        }
    }
    vector<vector<tuple<int, int, int>>> qry(n + 1);
    vector<int> ans(q);

    for (int i = 0; i < q; i++) {
        string s;
        cin >> s;
        int a = cal(s[0], s[1]);
        int b = cal(s[2], s[3]);
        int c = cal(s[4], s[5]);
        // cout << a << ' ' << b << ' ' << c << endl;
        qry[a].emplace_back(b, c, i);
    }

    priority_queue<Info> Q;
    for (int i = 1; i <= n; i++) {
        vector<int> d(n + 1, -1);
        Q.emplace(i, 0);
        while (!Q.empty()) {
            auto [u, mx] = Q.top();
            Q.pop();
            if (d[u] != -1) continue;  
            d[u] = mx;
            for (auto [v, w] : adj[u]) {
                Q.emplace(v, max(mx, w));
            }
        }       
        for (auto [b, c, idx] : qry[i]) {
            if (d[b] != -1 && d[b] <= c) {
                ans[idx] = 1;
            }
        }
    }

    for (int i = 0; i < q; i++) {
        cout << ans[i];
    }
    cout << "\n";

}   

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int t = 1;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3572kb

input:

2
5 2 4
0305040201
0404040404
030300
020500
050102
020501
6 2 4
030603010601
010203060504
030202
060402
050602
060401

output:

1011
0100

result:

ok 2 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 3648kb

input:

1
3 3 6
020202
030301
030201
020102
030203
010201
010303
020303
010202

output:

010101

result:

ok single line: '010101'

Test #3:

score: 0
Accepted
time: 68ms
memory: 4080kb

input:

200
10 10 5000
01060:04020305080709
0103070:060204050908
09070503080401060:02
050308010204090:0607
03010502040607080:09
03080109020504060:07
06050:09040302080107
07080305010409060:02
030809010:0204060507
0:060908070201050304
060700
090:03
09080:
070405
010703
0:0100
080601
030600
070206
0:0:09
08040...

output:

011110001101101111111111111111111101111111110111011110110110111011010111111111111111111101111111111110111111110111111111111101111111111110111111111111111111110001100111111111111111111111111011101111111111111111111111111111111111111111011011110100111110111111110111111100111111101110111111111101111110...

result:

ok 200 lines

Test #4:

score: 0
Accepted
time: 93ms
memory: 24700kb

input:

1
2000 1 1000000
M=:]A@8UAY7W2JJ4KEHIA[HSCQ1ENSC`JXR;F3PJ:_@41P9Z=9HR8P<<:DUXRR9^WOQFL?NZP6S@=J0^WE32=6;\U0?88]Q_RNPUMT6YU<4<S]H?:7OCQYOT4YAV1^764ENWSDBED>M7A:BI>KSIR48JQ9B=N\5T3N4A2aF0@>3TI81<G7;YE>W`NMP<:IT4CI3D0=GZC3I\CLQJQBA9BDIS9SAM55KaVA<Z@D=>:Y?CQHUQ5U3a6UVI8OKX9_FAF^7=5M85;<0;8YPAM<7Z7PP7A=N...

output:

000101000101100010001000000010010110000001000001001100000000010000100001000000101100000000010000001000000001110000010110100000111100100000001000000000011001010001000001001000100000000100011001100110010000010000101100000011111000000010000101010010000000010101000001010111100000100000000000000101000100...

result:

ok single line: '000101000101100010001000000010...0101000101000000000010101001000'

Test #5:

score: -100
Time Limit Exceeded

input:

1
2000 2000 1000000
FVAaH7GRPO;_11Da5J18@3SMG==\G8E8S^6:M4L0JH>MN5IXT>2<7WJ3U8LVJS=;;3F13>0D0>VOIIU@EPHG6ABL6;K?T1PXDERLG07]5C9^GDKG<SBMIW;`4W8P3=469TIPKH0O34523_J5C2MJ17D25Z@=K8H@M>WK<CMK7EO]BPD7B6AW741J5YIHIa1:ERSG>L3N2^F3<4F`DLE@2AA5=8GZK6:192FB736[WMV7:^DA2C:<LK040VJBM3M]WXU50`407TR_?PLF@5VL7OSL...

output:


result: