重要意义：复习好久没写的邻接表了。
Network, Seoul 2007, LA3902
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 Swhich 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 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 , find 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 {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 (3n
1,
000) which is the number of nodes of the tree network. The next line contains two integers s (1s
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
1.以S为根建树
2.从最底层的叶子开始思考(受其他因素影响小)，发现必须要在 最底层的 k*father上安装。
3.用一个叶子表来保存叶子节点，方便选择，否则会超时；
白书题解：
接下来，我们考虑深度最大的结点。比如结点8，应该在哪里放新的服务器来覆盖（“覆盖”一个叶子是指到该叶子的距离不超过k）它呢？只有结点5和结点4满足条件。显然，结点4比结点5划算，因为结点5所覆盖的叶子（6, 7, 8）都能被结点4所覆盖。一般的，对于深度最大的结点u，选择u的k级祖先是最划算的（父亲是1级祖先，父亲的父亲是2级祖先，以此类推）。证明过程留给读者自行思考。
下面给出上述算法的一种实现方法：每放一个新服务器，进行一次DFS，覆盖与它距离不超过k的所有结点。注意，本题只需要覆盖叶子，而不需要覆盖中间结点，而且深度不超过k的叶子已经被原始服务器覆盖，所以我们只需要处理深度大于k的叶结点即可。为了让程序更简单，我们可用nodes表避开“按深度排序”的操作
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#define uns unsigned
#define int64 long long
#ifdef WIN32
#define fmt64 "%I64d"
#else
#define fmt64 "%lld"
#endif
#define oo 0x13131313
#define FOR(i,l,r) for(int i=(l);i<=(r);i++)
#define ROF(i,r,l) for(int i=(r);i>=(l);i--)
#define maxn 1010
using namespace std;
struct edge { int to; edge *next; };
struct node { int ok;int deep;int father; edge *first;};
edge E[maxn*3],*EE=E+1;
node TREE[maxn];
int s,k,n,ynode;
struct YEZI { int num;int deep;};
YEZI yezi[maxn];
int numyezi=0;
void LINK(int u,int v)
{
*EE=(edge) {v,TREE[u].first},TREE[u].first=EE++;
*EE=(edge) {u,TREE[v].first},TREE[v].first=EE++;
}
int inputtree()
{
memset(TREE,0,sizeof(TREE));
memset(E,0,sizeof(E));
memset(yezi,0,sizeof(yezi));
EE=E+1;ynode=0;numyezi=0;
int u,v;
scanf("%d",&n);
scanf("%d%d",&s,&k);
TREE[s].ok=1;
FOR(i,1,n-1)
{
scanf("%d%d",&u,&v);
LINK(u,v);
}
}
void deepdfs(int now,int father,int deep)
{
edge *mm=TREE[now].first;
TREE[now].deep=deep;
TREE[now].father=father;
if(mm->next==NULL) { ynode++;yezi[++numyezi].num=now;yezi[numyezi].deep=deep;}
for(edge *p=TREE[now].first;p;p=p->next)
{
if(p->to!=father)
deepdfs(p->to,now,deep+1);
}
}
int markdfs(int now,int father,int deep)
{
edge *mm=TREE[now].first;
if(mm->next==NULL&&TREE[now].ok==0)
{
ynode--;
TREE[now].ok=1;
}
if(deep==k) return 0;
for(edge *p=TREE[now].first;p;p=p->next)
{
if(p->to!=father)
markdfs(p->to,now,deep+1);
}
return 0;
}
int cmp(const void *i,const void *j)
{
YEZI *ii=(YEZI *)i,*jj=(YEZI *)j;
return jj->deep-ii->deep;
}
int main()
{
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
int T,max=-1,maxnode=0,ans=0,t;
scanf("%d",&T);
while(T--)
{
ans=0;
inputtree();
deepdfs(s,-1,1);
markdfs(s,-1,0);
qsort(yezi+1,numyezi,sizeof(yezi[1]),cmp);
FOR(i,1,numyezi)
if(TREE[yezi[i].num].ok==0)
{
t=yezi[i].num;
FOR(i,1,k)
markdfs(t,-1,0);
ans++;
}
printf("%d\n",ans);
}
return 0;
}