Description
Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies.
On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of npoints (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges).
To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!
The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.
Print one integer – the number of lifelines in the tree.
4
1 2
1 3
1 4
3
5
1 2
2 3
3 4
3 5
4
In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
我们可以dfs遍历，但只遍历两个点就return，每个点遍历一次，就是/2就好啦
#include<bits/stdc++.h>
using namespace std;
vector<int>q[100000];
int flag[100000];
int ans;
void dfs(int v,int sum)
{
if(sum==2)
{
ans++;
return;
}
if(flag[v])
{
return;
}
flag[v]=1;
for(int i=0;i<q[v].size();i++)
{
if(!flag[q[v][i]])
{
dfs(q[v][i],sum+1);
}
}
}
int main()
{
int n;
int v,u;
ans=0;
cin>>n;
for(int i=1;i<=n-1;i++)
{
cin>>v>>u;
q[v].push_back(u);
q[u].push_back(v);
}
for(int i=1;i<=n;i++)
{
memset(flag,0,sizeof(flag));
dfs(i,0);
}
cout<<ans/2<<endl;
return 0;
}