Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 35091		Accepted: 11718

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.

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.

For each test case output the answer on a single line.

5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

8

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long LL;
const int maxn=2e4+5;
const int INF=1e9+7;
int cnt=0;
int head[maxn<<1];
int sonnum[maxn<<1],sonmax[maxn<<1];
int mi,pos;
int N,K;
int sum=0;
bool vis[maxn<<1];
LL ans;
vector<int>dis;
struct Edge
{int next,to,w;
}e[maxn<<1];
void Init()
{ans=0;cnt=0;for(int i=0;i<maxn;i++){head[i]=-1;sonnum[i]=sonmax[i]=0;vis[i]=false;}
}
void add_edge(int u,int v,int w)
{e[++cnt].to=v;e[cnt].w=w;e[cnt].next=head[u];head[u]=cnt;
}
void Query_size(int root,int pre)//求当前树的子树大小
{sum++;//存树的大小sonnum[root]=1;sonmax[root]=0;//注意初始化 for(int i=head[root];i!=-1;i=e[i].next){int v=e[i].to;if(vis[v]||v==pre) continue;Query_size(v,root);sonnum[root]+=sonnum[v];//子树节点有多少个sonmax[root]=max(sonmax[root],sonnum[v]);//最大的子树节点个数
    }
}
void Query_root(int root,int pre,int sum)//求当前树的重心
{for(int i=head[root];i!=-1;i=e[i].next){int v=e[i].to;if(vis[v]||v==pre) continue;Query_root(v,root,sum);}int ma=max(sonmax[root],sum-sonnum[root]);if(mi>ma){mi=ma;pos=root;}
}
void Query_dis(int root,int pre,int d)
{dis.push_back(d);for(int i=head[root];i!=-1;i=e[i].next){int v=e[i].to,w=e[i].w;if(vis[v]||v==pre) continue;Query_dis(v,root,d+w);}
}
int cal(int root,int d)
{int ret=0;dis.clear();//存所有子节点到本身的距离Query_dis(root,0,d);sort(dis.begin(),dis.end());int i=0,j=dis.size()-1;while(i<j){while(i<j&&dis[i]+dis[j]>K) j--;ret+=j-i;i++;}return ret;
}
void dfs(int root,int pre)
{sum=0;mi=INF;Query_size(root,pre);Query_root(root,pre,sum);//
    int rt=pos;ans+=cal(rt,0);//pos为找到的重心vis[rt]=true;//一定要标记 否则会往回走for(int i=head[rt];i!=-1;i=e[i].next){int v=e[i].to,w=e[i].w;if(vis[v]) continue;ans-=cal(v,w);dfs(v,rt);}
}
int main()
{while(scanf("%d%d",&N,&K)!=EOF){Init();if(N==0&&K==0) break;for(int i=1;i<N;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);add_edge(u,v,w);add_edge(v,u,w);}dfs(1,0);printf("%lld\n",ans);}return 0;
}