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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#571776 | #9310. Permutation Counting 4 | RFD | TL | 3ms | 11472kb | C++14 | 1.1kb | 2024-09-18 08:29:53 | 2024-09-18 08:29:53 |
Judging History
answer
#include <iostream>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <unordered_set>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;//开long long,求求你了,记得开long long
const LL N=1e6+10;
const LL INF=1e18;
const double small=1e-16;
//千万不要用puts()和gets(),求求你了
LL n;
int h[N];
int e[2*N],ne[2*N],idx;
unordered_set<int> has;
bool st[N];
int ans=0;
void add(int x,int y)
{
e[idx]=y;
ne[idx]=h[x];
h[x]=idx++;
}
void dfs(int u)
{
if(st[u]==0)
{
ans++;
st[u]=1;
}
for(int i=h[u];i!=-1;i=ne[i])
{
int k=e[i];
if(!st[k])
{
dfs(k);
}
}
}
void solve()
{
cin>>n;
memset(h,-1,sizeof h);
memset(st,false,sizeof st);
ans=0;
idx=0;
for(int i=1;i<=n;i++)
{
int x,y;
cin>>x>>y;
add(x-1,y);
add(y,x-1);
has.insert(x-1);
has.insert(y);
}
dfs(0);
if(ans==n+1)cout<<"1"<<endl;
else cout<<"0"<<endl;
}
int main()
{
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 3ms
memory: 11472kb
input:
4 5 1 2 1 5 1 2 1 2 2 2 5 1 1 2 4 2 3 5 5 3 4 5 3 5 1 2 3 4 3 5 3 3 5 1 5 1 4 4 5 5 5 1 2
output:
0 1 0 0
result:
ok 4 tokens
Test #2:
score: -100
Time Limit Exceeded
input:
66725 14 7 7 4 6 7 8 8 13 2 13 6 13 6 10 14 14 1 10 9 11 7 9 3 8 4 12 5 12 12 2 6 3 6 7 11 2 5 1 1 6 12 8 12 2 3 7 9 7 8 1 10 1 4 10 4 8 4 4 6 10 9 10 2 3 2 7 1 3 3 4 2 2 3 10 20 3 12 10 14 19 20 19 20 1 9 7 9 13 16 17 17 16 18 2 11 5 19 6 17 11 17 3 6 3 11 7 20 8 17 3 18 10 15 9 20 16 5 10 2 10 2 1...
output:
1 1 0 0 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 0 0 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 0 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1 1 ...