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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
#571776#9310. Permutation Counting 4RFDTL 3ms11472kbC++141.1kb2024-09-18 08:29:532024-09-18 08:29:53

Judging History

你现在查看的是最新测评结果

  • [2024-09-18 14:56:40]
  • hack成功,自动添加数据
  • (/hack/835)
  • [2024-09-18 14:41:06]
  • hack成功,自动添加数据
  • (/hack/831)
  • [2024-09-18 08:29:53]
  • 评测
  • 测评结果:TL
  • 用时:3ms
  • 内存:11472kb
  • [2024-09-18 08:29:53]
  • 提交

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;
}

详细

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
...

result: