UVA

A network is composed of N computers connected by N - 1 communication links such that any two computers can be communicated via a unique route. Two computers are said to be adjacent if
there is a communication link between them. The neighbors of a computer is the set of computers which are adjacent to it. In order to quickly access and retrieve large amounts of information, we need to select some computers acting
as servers to provide resources to their neighbors. Note that a server can serve all its neighbors. A set of servers in the network forms a perfect service if every client (non-server) is served by exactly
one server. The problem is to find a minimum number of servers which forms a perfect service, and we call this number perfect service number.

We assume that N(10000) is a positive integer and these N computers
are numbered from 1 to N . For example, Figure 1 illustrates a network comprised of six computers, where black nodes represent servers and white nodes represent clients. In Figure 1(a), servers 3 and 5 do not form a perfect
service because client 4 is adjacent to both servers 3 and 5 and thus it is served by two servers which contradicts the assumption. Conversely, servers 3 and 4 form a perfect service as shown in Figure 1(b). This set also has the minimum cardinality. Therefore,
the perfect service number of this example equals two.

Your task is to write a program to compute the perfect service number.

Input

The input consists of a number of test cases. The format of each test case is as follows: The first line contains one positive integer, N , which represents the number of computers in the network. The next N -
1 lines contain all of the communication links and one line for each link. Each line is represented by two positive integers separated by a single space. Finally, a ` 0' at the (N + 1) -th line indicates the
end of the first test case.

The next test case starts after the previous ending symbol `0'. A `-1' indicates the end of the whole inputs.

Output

The output contains one line for each test case. Each line contains a positive integer, which is the perfect service number.

Sample Input

6
1 3
2 3
3 4
4 5
4 6
0
2
1 2
-1

Sample Output

2

1

这题难在建立状态

这里面有三种状态的点 1该点是服务器,2该点不是服务器但是父亲节点是(这样的话子节点不能是服务器),3该点不是服务器父亲节点也不是(这样的话子节点必须要有一个是服务器)

状态建好了,转移并不难

dp[u][0]={min(dp[son[u]][0],dp[son[u]][1])}+1;//表示该点是服务器

dp[u][1]={dp[son[u]][2]}     //dp[u][1]表示该点不是服务器,父亲节点是服务器

dp[u][2]={dp[son[u]][2]} +min(-dp[son[u]][2],dp[son[u]][0]) //他是由子节点只有一个是服务器转移来的

//
// Created by Zeroxf on 2015-08-09-20.06
// Copyright: (c) 2015 Zeroxf. All rights reserved
//
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<stack>
#include<map>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 1e4+10;
const int INF  = 1e9;
vector <int> G[maxn] ,vertices;
int p[maxn], dp[maxn][3],n;
void dfs(int u,int fa){
    p[u] = fa;
    vertices.push_back(u);
    for( int i =0; i < G[u].size() ; i++){
        int v = G[u][i];
        if(v!=p[u]) dfs(v,u);
    }
}
int main(){
#ifdef LOCAL
    freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
 #endif
    int a,b;
    while(cin>>n&&n){
        memset(p, 0 ,sizeof p);
        memset(dp, 0,sizeof dp);
        for(int i=0;i<maxn;i++) G[i].clear();
        for( int i =0;i<n-1;i++){
            cin>>a>>b;
            a--;b--;
            G[a].push_back(b);
            G[b].push_back(a);
        }
        dfs(0,-1);
        for(int i = vertices.size()-1; i>=0; i--){
            int u = vertices[i];
            dp[u][0] = 1;dp[u][1] = 0;
            for (int j = 0; j< G[u].size(); j++){
                int v = G[u][j];
                if(v==p[u]) continue;
                dp[u][1] += dp[v][2];
                dp[u][0] += min(dp[v][0],dp[v][1]);
                if(dp[u][1]>=INF) dp[u][1] = INF;
                if(dp[u][0]>= INF) dp[u][0] = INF;
            }
            dp[u][2] = INF;
            for(int j=0;j<G[u].size();j++){
                int v = G[u][j];
                if(v==p[u]) continue;
                dp[u][2] = min (dp[u][1]+dp[v][0]-dp[v][2],dp[u][2]);
            }
        }
        cout<<min(dp[0][0],dp[0][2])<<endl;
        cin>>n;
    }
    return 0;
}