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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
#693850#7738. Equivalent RewritingRosmontis_L#RE 1ms3568kbC++201.6kb2024-10-31 16:50:242024-10-31 16:50:41

Judging History

你现在查看的是最新测评结果

  • [2024-10-31 16:50:41]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:3568kb
  • [2024-10-31 16:50:24]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
const ll N = 100010;
vector<ll> edge[N], p[N];
ll teg[N];

void solve(){
    ll n, m;
    cin >> n >> m;
    for(int i = 1; i <= n; i ++){
        edge[i].clear();
        p[i].clear();
        teg[i] = 0;
    } 

    for(int i = 1; i <= n; i ++){
        ll siz;
        cin >> siz;
        for(int j = 1; j <= siz; j ++){
            ll x;
            cin >> x;
            p[x].push_back(i);
        }
    }

    for(int i = 1; i <= n; i ++){
        for(int j = 0; j < p[i].size() - 1; j ++){
            edge[p[i][j]].push_back(p[i].back());
            teg[p[i].back()] ++;
        }
    }
    vector<ll> ans(n + 10);
    ll sum = 0, now = 0;
    for(int i = n; i >= 1; i --){
        if(teg[i] != 0) continue;
        priority_queue<ll> q;
        q.push(i);
        while(q.size()){
            auto u = q.top();
            q.pop();
            ans[++ now] = u;
            for(auto v : edge[u]){
                if(-- teg[v] == 0) q.push(v);
            }
        }
    }
    bool ok = 1;
    for(int i = 1; i <= n; i ++){
        if(ans[i] != i){
            ok = 0;
            break;
        }
    }
    if(ok){
        cout << "No\n";
        return;
    }
    cout << "Yes\n";
    for(int i = 1; i <= n; i ++){
        cout << ans[i] << " \n"[i == n];
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    int t = 1;
    cin >> t;
    while(t --){
        solve();
    }
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3568kb

input:

3
3 6
3 3 1 5
2 5 3
2 2 6
2 3
3 1 3 2
2 3 1
1 3
2 2 1

output:

Yes
3 1 2
No
No

result:

ok OK. (3 test cases)

Test #2:

score: -100
Runtime Error

input:

1
10 5
2 2 4
4 1 3 4 2
1 2
3 2 1 4
4 5 2 4 3
3 2 5 4
3 5 4 2
3 1 3 2
5 1 4 2 3 5
1 4

output:


result: