ACM: uva 1276 - Network

Network

Consider a tree network
with n nodes where the
internal nodes correspond to servers and the terminal nodes
correspond to clients. The nodes are numbered from 1
to n . Among the servers,
there is an original
server S which provides
VOD (Video On Demand) service. To ensure the quality of service for
the clients, the distance from each client to the VOD
server S should not exceed
a certain value k . The
distance from a node u to
a node v in the tree is
defined to be the number of edges on the path
fromu to v .
If there is a nonempty
subset C of clients such
that the distance from
eachu in C to S is
greater than k , then
replicas of the VOD system have to be placed in some servers so
that the distance from each client to the nearest VOD server (the
original VOD system or its replica)
is k or less.
Given a tree network, a
server S which has VOD
system, and a positive
integer k , find the
minimum number of replicas necessary so that each client is within
distancek from the nearest server which has
the original VOD system or its replica.
For example, consider the following tree
network.
 
In the above tree, the set of clients is {1, 6, 7,
8, 9, 10, 11, 13}, the set of servers is {2, 3, 4, 5, 12, 14}, and
the original VOD server is located at node 12.
For k =
2 , the quality of service is not guaranteed with
one VOD server at node 12 because the clients in {6, 7, 8, 9, 10}
are away from VOD server at
distance > k .
Therefore, we need one or more replicas. When one replica is placed
at node 4, the distance from each client to the nearest server of
{12, 4} is less than or equal to 2. The minimum number of the
needed replicas is one for this example.
Input 
Your program is to read the input from standard
input. The input consists
of T test cases. The
number of test cases (T ) is given in the
first line of the input. The first line of each test case contains
an integer n (3 n
1, 000) which is the number of nodes of the tree
network. The next line contains two
integers s (1 s
n)and k (k
1) where s is
the VOD server and k is
the distance value for ensuring the quality of service. In the
following n - 1 lines,
each line contains a pair of nodes which represent an edge of the
tree network.
Output 
Your program is to write to standard output. Print
exactly one line for each test case. The line should contain an
integer that is the minimum number of the needed replicas.
Sample
Input 
2 14

12 2

1 2

2 3

3 4

4 5

5 6

7 5

8 5

4 9

10 3

2 12

12 14

13 14

14 11

14

3 4

1 2

2 3

3 4

4 5

5 6

7 5

8 5

4 9

10 3

2 12

12 14

13 14

14 11

Sample Output 
1
0
 

题意: 用最少的服务器覆盖全部的节点, 每个服务器的覆盖范围是k, 计算最少还安装需要多少个服务器.

 

解题思路:

  1. 先将无根树转换成有根树,
原始服务器作为根(root).

  2. 问题变成需要覆盖全部的叶子节点即可,
采用"最远覆盖"这样解是最优的(将服务器覆盖范围完全利用).

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <vector>

using namespace std;

#define MAX 1005

int n, s, k;

vector g[MAX], node[MAX];

int fa[MAX];

bool flag[MAX];

void dfs(int u, int f, int d)

{

 fa[u] = f;

 int size = g[u].size();

 if(d > k
&& size == 1)
node[d].push_back(u);

 for(int i = 0; i < size;
++i)

 {

  int v = g[u][i];

  if(v != f) dfs(v, u,
d+1);

 }

}

void dfs1(int u, int f, int d)

{

 flag[u] = true;

 int size = g[u].size();

 for(int i = 0; i < size;
++i)

 {

  int v = g[u][i];

  if(v != f
&& d < k) dfs1(v, u,
d+1);

 }

}

int solve()

{

 int result = 0;

 memset(flag, false, sizeof(flag));

 for(int d = n-1; d > k; --d)

 {

  int size =
node[d].size();

  for(int i = 0; i
< size; ++i)

  {

   int u =
node[d][i];

   if(flag[u])
continue;

   int v =
u;

   for(int j =
0; j < k; ++j) v = fa[v];

   dfs1(v, -1,
0);

   result++;

  }

 }

 return result;

}

int main()

{

// freopen("input.txt", "r", stdin);

 int caseNum, i;

 scanf("%d", &caseNum);

 while(caseNum--)

 {

  scanf("%d %d %d",
&n, &s, &k);

  for(i = 1; i <=
n; ++i)

  {

   g[i].clear();

   node[i].clear();

  }

  

  for(i = 0; i <
n-1; ++i)

  {

   int u,
v;

   scanf("%d
%d", &u, &v);

   g[u].push_back(v);

   g[v].push_back(u);

  }

  

  dfs(s, -1, 0);

  printf("%d\n", solve());

 }

 return 0;

}