Description
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0
Sample Output
8
Source
树的重心：树上有一个结点,其所有的子树中最大的子树节点数最少,这个点就是这棵树的重心
点分治：先找出树的重心作为树根，统计其子树的答案，然后递归处理各子树，并去除重复算的答案。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 using namespace std; 7 const int mxn=30200; 8 //base 9 int n,k; 10 int ans; 11 //node 12 struct edge{ 13 int nx,to; 14 int w; 15 }e[40020]; 16 int head[mxn],size[mxn],mc[mxn],dis[mxn];//邻接表队首 点的子树尺寸 点重量 结点深度 17 int cnt; 18 //tree 19 int root,mn=mxn; 20 int tot; 21 bool vis[mxn]; 22 // 23 void add_edge(int u,int v,int len){ 24 e[++cnt]=(edge){head[u],v,len};head[u]=cnt; 25 e[++cnt]=(edge){head[v],u,len};head[v]=cnt; 26 } 27 void dfssize(int u,int fa){//求结点子树尺寸 28 size[u]=1; 29 mc[u]=0; 30 for(int i=head[u];i;i=e[i].nx){ 31 int v=e[i].to; 32 if(v!=fa && !vis[v]){ 33 dfssize(v,u); 34 size[u]+=size[v]; 35 mc[u]=size[v]>mc[u]? size[v]:mc[u]; 36 } 37 } 38 return; 39 } 40 void findmc(int u,int fa,int rt){//求树的重心 41 mc[u]=max(mc[u],size[rt]-size[u]); 42 if(mc[u]<mn){mn=mc[u];root=u;}//符合要求则更新树根 43 for(int i=head[u];i;i=e[i].nx){ 44 int v=e[i].to; 45 if(v!=fa && !vis[v])findmc(v,u,rt); 46 } 47 return; 48 } 49 void dist(int u,int fa,int d){//求u的子树的结点深度 50 dis[++tot]=d; 51 for(int i=head[u];i;i=e[i].nx){ 52 int v=e[i].to; 53 if(v!=fa && !vis[v]) 54 dist(v,u,d+e[i].w); 55 } 56 } 57 int calc(int x,int d){//计算以x为中心，所有经过x的路径上符合要求的点对的数量 58 int res=0; 59 tot=0; 60 dist(x,0,d);//计算x为根的树的各结点深度 61 sort(dis+1,dis+tot+1);//深度由小到大排序 62 int l=1,r=tot; 63 while(l<r){ 64 while(dis[l]+dis[r]>k && l<r )r--;//两结点深度相加就是距离 65 res+=r-l; 66 l++; 67 } 68 return res; 69 } 70 71 void dfs(int rt){//处理以rt为根的子树 72 // printf("test!: %d %d ",rt,root); 73 mn=n;//记得 74 dfssize(rt,0); 75 findmc(rt,0,rt); 76 ans+=calc(root,0);//在以rt为根的子树中找到重心后，以其为中心计算符合要求的点对数 77 vis[root]=1; 78 for(int i=head[root];i;i=e[i].nx){ 79 int v=e[i].to; 80 if(!vis[v]){ 81 ans-=calc(v,e[i].w);//减去以v为中心的答案数，避免之后计算重复 82 dfs(v); 83 } 84 } 85 } 86 int main(){ 87 while(scanf("%d",&n) && n){ 88 memset(vis,0,sizeof(vis)); 89 memset(head,0,sizeof(head)); 90 ans=0; 91 cnt=0;//初始化！初始化！ 92 scanf("%d",&k); 93 for(int i=1,u,v,d;i<n;i++){ 94 scanf("%d%d%d",&u,&v,&d); 95 add_edge(u,v,d); 96 } 97 dfs(1);//随便找个结点开始计算 98 printf("%d ",ans); 99 } 100 return 0; 101 }