树形DP
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Perfect Service
Description 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. 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
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 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 Source |
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/* ***********************************************
Author :CKboss
Created Time :2015年02月22日 星期日 12时09分02秒
File Name :POJ3398_2.cpp
************************************************ */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int maxn=11000;
const int INF=1e6;
struct Edge
{
int to,next;
}edge[maxn*2];
int Adj[maxn],Size;
int dp[maxn][3],n;
void init()
{
Size=0;
for(int i=0;i<=n+10;i++)
{
Adj[i]=-1;
dp[i][0]=1; dp[i][1]=0; dp[i][2]=INF;
}
}
void add_edge(int u,int v)
{
edge[Size].to=v;
edge[Size].next=Adj[u];
Adj[u]=Size++;
}
void dfs(int u,int fa)
{
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(v==fa) continue;
dfs(v,u);
dp[u][0]+=min(dp[v][0],dp[v][1]);
dp[u][1]+=dp[v][2];
dp[u][2]=min(dp[u][2],dp[v][0]-dp[v][2]);
}
dp[u][2]+=dp[u][1];
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d",&n)!=EOF)
{
if(n==-1) break;
else if(n==0) continue;
init();
for(int i=0;i<n-1;i++)
{
int a,b;
scanf("%d%d",&a,&b);
add_edge(a,b); add_edge(b,a);
}
dfs(1,0);
printf("%d\n",min(dp[1][0],dp[1][2]));
}
return 0;
}