HUD5423 Rikka with Tree(DFS)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5423
Rikka with Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 321 Accepted Submission(s): 162
For a tree T
, let F(T,i)
be the distance between vertice 1 and vertice i
.(The length of each edge is 1).
Two trees A
and B
are similiar if and only if the have same number of vertices and for each i
meet F(A,i)=F(B,i)
.
Two trees A
and B
are different if and only if they have different numbers of vertices or there exist an number i
which vertice i
have different fathers in tree A
and tree B
when vertice 1 is root.
Tree A
is special if and only if there doesn't exist an tree B
which A
and B
are different and A
and B
are similiar.
Now he wants to know if a tree is special.
It is too difficult for Rikka. Can you help her?
For each testcase, the first line contains a number n(1≤n≤1000)
.
Then n−1
lines follow. Each line contains two numbers u,v(1≤u,v≤n)
, which means there is an edge between u
and v
.
2 3
2 3
1 4
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std; bool ok;
vector<int> a[];
int num[]; void dfs(int u, int fa, int d)
{
num[d]++;
int len = a[u].size();
for(int i=; i<len; i++)
{
if(a[u][i]==fa) continue;
dfs(a[u][i], u, d+);
}
} int main()
{
int n;
while(~scanf("%d", &n))
{
for(int i=; i<; i++)
a[i].clear();
memset(num, , sizeof(num));
for(int i=; i<n; i++)
{
int x, y;
scanf("%d%d", &x, &y);
a[x].push_back(y);
a[y].push_back(x);
}
ok = false;
dfs(, -, );
for(int i=; i<; i++)
{
if(num[i-]>&&num[i])
ok = true;
}
if(ok) printf("NO\n");
else printf("YES\n");
}
return ;
}
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