Problem Description

Farmer John has decided to give each of his cows a cell phone in hopes to encourage their social interaction. This, however, requires him to set up cell phone towers on his N (1 ≤ N ≤ 10,000) pastures (conveniently numbered 1..N) so they can all communicate.

Exactly N-1 pairs of pastures are adjacent, and for any two pastures A

 and B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B) there is a sequence of adjacent pastures such that A is the first pasture in the sequence and B is the last. Farmer John can only place cell phone towers in the pastures, and each tower has enough range to provide service to the pasture it is on and all pastures adjacent to the pasture with the cell tower.

Help him determine the minimum number of towers he must install to provide cell phone service to each pasture.

Input

<p>* Line 1: A single integer: <i>N</i><br>* Lines 2..<i>N</i>: Each line specifies a pair of adjacent pastures with two space-separated integers: <i>A</i> and <i>B</i></p>

Output

<p>* Line 1: A single integer indicating the minimum number of towers to install</p>

Sample Input

5 1 3 5 2 4 3 3 5

Sample Output

2

n個點，農場主準備給每個點的牛一個電話，如果在某點設立電線杆，那麼其相連的點和當前點都可以互相通電話，請問最少設立多少個電線杆才能使得所有點的牛都可以互相打電話。

樹的最小支配集，這裡用貪心

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=20020;
struct edge
{
    int v,next;
}edge[maxn];
bool vis[maxn];

int head[maxn];
int father[maxn];
int newpos[maxn];
int n,now,cnt;
void init()
{
    memset(head,-1,sizeof(head));
    cnt=0;
    memset(father,0,sizeof(father));
    memset(newpos,0,sizeof(newpos));
    memset(vis,0,sizeof(vis));
    memset(edge,0,sizeof(edge));
}
void addedge(int u,int v)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void dfs(int x)
{
    newpos[now++]=x;
    for(int i=head[x];i!=-1;i=edge[i].next)
    {
        if(!vis[edge[i].v])
        {vis[edge[i].v]=1;
        father[edge[i].v]=x;
        dfs(edge[i].v);

    }
    }
}
bool s[maxn],set[maxn];
int greedy()
{
    memset(s,0,sizeof(s));
    memset(set,0,sizeof(set));
    int ans=0;
    for(int i=n;i>=1;i--)
    {
        int t=newpos[i];
        if(!s[t])
        {
            if(!set[father[t]])
            {set[father[t]]=1;
            ans++;
            }

        s[t]=1;
        s[father[t]]=1;
        s[father[father[t]]]=1;
    }
    }

    return ans;
}
int main()
{while(~scanf("%d",&n))
{int u,v;
init();
for(int i=1;i<=n-1;i++)
{scanf("%d%d",&u,&v);

addedge(u,v);
addedge(v,u);}
now=0;

dfs(1);
printf("%d\n",greedy());

}
    return 0;
}