UVALive 3902 Network （树+dfs）
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 de ned to be the number of edges on the path from u
to v. If there is a nonempty subset C of clients such that the distance from each u 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, nd the minimum
number of replicas necessary so that each client is within distance k 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 f1, 6, 7, 8, 9, 10, 11, 13g, the set of servers is f2, 3, 4, 5, 12,
14g, 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 f6, 7, 8, 9, 10g 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
f12, 4g 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 rst line of the input. The rst 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
题目大意就是给你一棵树，它的根节点编号为s，现在根节点要给叶子结点发送信息。如果叶子节点与根结点的距离大于k则接收不到信息。
现为了让所有的节点都接受到信息，在节点上建立信息发送站，信息发送站也可以发送信息。问最少加几个信息发送站？
首先我们思考如果节点距离根节点的距离小于k，这样的节点我们是不用管的。剩下的怎么办呢?
我们建立一个vector g，g[i]里面存放距离根节点距离为i的点的编号,g里面的节点都是距离根节点距离大于k的点。
现在我们从离根节点最远的一群节点向上找k次父亲，让这个祖先变成发信站，然后从这个发信站向下dfs，将他所有的子孙覆盖。这样情况就是最优的了。
代码如下：
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define M 1050 4 int n,k,s; 5 int tot,father[M]; 6 vector <int>edge[M]; 7 vector <int>g[M]; 8 bool covered[M]; 9 void dfs (int u,int pre,int d) 10 { 11 father[u]=pre; 12 int sz=edge[u].size(),v; 13 if (sz==1&&d>k) 14 g[d].push_back(u); 15 for (int i=0;i<sz;++i) 16 { 17 v=edge[u][i]; 18 if (v==pre) 19 continue; 20 dfs(v,u,d+1); 21 } 22 } 23 void dfs2(int u,int pre,int d) 24 { 25 covered[u]=true; 26 int v=0; 27 for (int i=0;i<edge[u].size();++i) 28 { 29 v=edge[u][i]; 30 if(v==pre) 31 continue; 32 if (d<k) 33 dfs2(v,u,d+1); 34 } 35 } 36 int main() 37 { 38 int t; 39 //freopen("de.txt","r",stdin); 40 //freopen("in.txt","w",stdout); 41 scanf("%d",&t); 42 while (t--) 43 { 44 scanf("%d",&n); 45 scanf("%d%d",&s,&k); 46 memset(father,0,sizeof father); 47 memset(covered,0,sizeof covered); 48 int ans=0; 49 for (int i=0;i<=n;++i) 50 { 51 edge[i].clear(); 52 g[i].clear(); 53 } 54 for (int i=1;i<n;++i) 55 { 56 int u,v; 57 scanf("%d%d",&u,&v); 58 edge[u].push_back(v); 59 edge[v].push_back(u); 60 } 61 dfs(s,-1,0); 62 for (int i=n-1;i>k;--i) 63 { 64 for (int j=0;j<g[i].size();++j) 65 { 66 int u=g[i][j]; 67 if (covered[u]) 68 continue; 69 int v=u; 70 for (int j=0;j<k;++j) 71 v=father[v]; 72 dfs2(v,-1,0); 73 ans++; 74 } 75 } 76 printf("%d\n",ans); 77 } 78 return 0; 79 }
