Codeforces 690F1 - Tree of Life (easy)

F1. Tree of Life (easy)

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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
n points (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!

Input
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.

Output
Print one integer – the number of lifelines in the tree.

Examples

Input

4
1 2
1 3
1 4

Output

3

Input

5
1 2
2 3
3 4
3 5

Output

4

Note
In the second sample, there are four lifelines: paths between vertices
1 and 3, 2 and
4, 2 and
5, and 4 and 5.

题目大意就是给一棵树让你求有多少组长度为2(两点之间长度为1)的路.

就是用dfs搜一下,顺便复习一下邻接链表这种数据结构.

注意存边的数组开大点,不然会莫名其妙越界.....

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
struct p
{

int node,next;

}E[500005];
int head[10005];
int fre=0;
int n;
int ans;
int vis[10005];
void insertt(int x,int y)
{

E[fre].node=y;
E[fre].next=head[x];
head[x]=fre;
fre++;
}
void dfs(int star,int deep)
{
if (deep==2)
{
ans++;
return;
}
if (!vis[star])
for (int i=head[star];i!=-1;i=E[i].next)
{
if (!vis[E[i].node])
{
vis[star]=1;
dfs(E[i].node,deep+1);
}
}
}
int main()
{

scanf("%d",&n);
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
for (int i=1;i<n;i++)
{
int a,b;
scanf("%d%d",&a,&b);
insertt(a,b);
insertt(b,a);
}
ans=0;
for (int i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
dfs(i,0);
}
printf("%d",ans/2);
}