Perfect Service - UVa 1218 dp-优快云博客

本文探讨了在一棵树形网络中，如何确定最少数量的服务器来确保每个客户端仅由一个服务器提供服务的问题。通过使用深度优先搜索和动态规划的方法，文章提出了一种高效的解决方案。

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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 - 1lines 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

Sample Output

2 
1

题意：在一个树形的网络上，每个点是一个计算机，每个计算机可以当做服务器，一个计算机要么是服务器，要么和他直接连的计算机有服务器，问至少需要多少个服务器。

思路：dp[i][0 1 2]记录三种状况，当他是服务器时，子树的最小权值；当他不是服务器，并且和他直接相连的有服务器时，子树的最小权值；当他不是服务器，并且和他直接相连的没有服务器时，子树的最小权值。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
vector <int> vc[10010];
int dp[10010][3],vis[10010],t=0,INF=100000000;
void dfs(int u)
{ int len=vc[u].size(),i,j;
  vis[u]=t;
  dp[u][0]=1;dp[u][1]=0;dp[u][2]=0;
  vector <int > sub;
  for(i=0;i<len;i++)
   if(vis[vc[u][i]]!=t)
   { dfs(vc[u][i]);
     sub.push_back(vc[u][i]);
   }
  bool flag=false;
  len=sub.size();
  if(len==0)
  { dp[u][1]=INF;
    return;
  }
  for(i=0;i<len;i++)
  { dp[u][0]+=min(dp[sub[i]][0],min(dp[sub[i]][1],dp[sub[i]][2]));
    dp[u][2]+=dp[sub[i]][1];
    dp[u][1]+=min(dp[sub[i]][0],dp[sub[i]][1]);
    if(dp[sub[i]][0]<=dp[sub[i]][1])
     flag=true;
  }
  if(!flag)
  { int k=dp[u][1];
    dp[u][1]=k-dp[sub[0]][1]+dp[sub[0]][0];
    for(i=1;i<len;i++)
     dp[u][1]=min(dp[u][1],k-dp[sub[i]][1]+dp[sub[i]][0]);
  }
}
int main()
{ int n,m,i,j,k,u,v;
  while(true)
  { t++;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
     vc[i].clear();
    for(i=1;i<n;i++)
    { scanf("%d%d",&u,&v);
      vc[u].push_back(v);
      vc[v].push_back(u);
    }
    dfs(1);
    printf("%d\n",min(dp[1][0],dp[1][1]));
    scanf("%d",&k);
    if(k==-1)
     break;
  }
}