Problem Description
Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses, and all houses are connected together. To make the race more interesting, he requires that every participant must start from a different house and run AS FAR AS POSSIBLE without passing a road more than once. The distance difference between the one who runs the longest distance and the one who runs the shortest distance is called “race difference” by Bob. Bob does not want the “race difference”to be more than Q. The houses are numbered from 1 to N. Bob wants that the No. of all starting house must be consecutive. He is now asking you for help. He wants to know the maximum number of starting houses he can choose, by other words, the maximum number of people who can take part in his race.
Input
There are several test cases.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The following N-1 lines, each contains three integers, x, y and z, indicating that there is a road of length z connecting house x and house y.
The following M lines are the queries. Each line contains an integer Q, asking that at most how many people can take part in Bob’s race according to the above mentioned rules and under the condition that the“race difference”is no more than Q.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The following N-1 lines, each contains three integers, x, y and z, indicating that there is a road of length z connecting house x and house y.
The following M lines are the queries. Each line contains an integer Q, asking that at most how many people can take part in Bob’s race according to the above mentioned rules and under the condition that the“race difference”is no more than Q.
The input ends with N = 0 and M = 0.
(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)
Output
For each test case, you should output the answer in a line for each query.
Sample Input
5 5
1 2 3
2 3 4
4 5 3
3 4 2
1
2
3
4
5
0 0
Sample Output
1
3
3
3
5
题意：有一颗树有n个节点编号从1~n，这n-1条边都有一个边权表示两点之间的距离，现在要求出从每个点出发不重复经过同一条边能到达的最远距离，然后m次询问，每次输入一个Q，求从连续编号的起点出发达到最远距离之间的最大值和最小值之差小于等于Q的最大区间长？
思路：用前向星建树（定1号节点为根节点），一次搜索求出所有节点 以当前节点为根节点时到达子树叶子节点的最远距离和次远距离，然后再进行一次搜索求出每个节点能到达的最远距离（可能是到孩子叶子节点，也可能是经过父亲节点而来的叶子节点），最后用RMQ中的ST算法配合尺取求出满足条件的最大区间长。
注意：使用ST算法，查询时不要使用函数log2（x），这个很费时间，自己求就行；另外可以优化一下尺取。
代码如下：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const int N=;
int h[N],H[N],H2[N],ans[N],tot;
int f1[N][],f2[N][];
struct edge
{
int to;
int next;
int x;
}e[*N]; void add(int u,int v,int x)
{
e[tot].to=v;
e[tot].x=x;
e[tot].next=h[u];
h[u]=tot++; e[tot].to=u;
e[tot].x=x;
e[tot].next=h[v];
h[v]=tot++;
}
void getH(int id,int fa)
{
H[id]=;
H2[id]=;
for(int i=h[id];i!=-;i=e[i].next)
{
if(e[i].to==fa) continue;
getH(e[i].to,id);
int tmp=H[e[i].to]+e[i].x;
if(H[id]<tmp)
{
H2[id]=H[id];
H[id]=tmp;
}
else if(H2[id]<tmp) H2[id]=tmp;
}
}
void dfs(int id,int fa,int deep)
{
ans[id]=max(deep,H[id]);
for(int i=h[id];i!=-;i=e[i].next)
{
if(e[i].to==fa) continue;
int tmp=deep;
int d=H[e[i].to]+e[i].x;
if(d==H[id]) tmp=max(tmp,H2[id]);
else tmp=max(tmp,H[id]);
dfs(e[i].to,id,tmp+e[i].x);
}
}
void cal(int n)
{
for(int i=;i<=n;i++) f1[i][]=ans[i],f2[i][]=ans[i];
int len=;
for(int s=;len<=n;s++,len*=)
{
for(int i=;i+len-<=n;i++)
{
f1[i][s]=max(f1[i][s-],f1[i+len/][s-]);
f2[i][s]=min(f2[i][s-],f2[i+len/][s-]);
}
}
}
int get(int i,int j)
{
int len=-;
int t=j-i+;
while(t)
{
len++;
t>>=;
}
return max(f1[i][len],f1[j-(<<len)+][len])-min(f2[i][len],f2[j-(<<len)+][len]);
} void init()
{
tot=;
memset(h,-,sizeof(h));
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)&&(n+m))
{
init();
for(int i=;i<n;i++)
{
int u,v,x;
scanf("%d%d%d",&u,&v,&x);
add(u,v,x);
}
getH(,-);
dfs(,-,);
cal(n);
while(m--)
{
int Q; scanf("%d",&Q);
int res=;
int pos=;
for(int i=;i<=n;i++)
{
if(i-pos+<=res) continue; ///优化一下;
int tmp=get(pos,i);
while(tmp>Q)
{
pos++;
tmp=get(pos,i);
}
res=max(res,i-pos+);
}
printf("%d\n",res);
}
}
return ;
}
/**
4 6
1 2 2
1 3 4
3 4 3
*/ /**
18 7984
1 2 2
1 3 4
1 4 3
1 5 5 2 6 1
2 7 3
2 8 7
3 9 2
3 10 4
4 11 2
4 12 2
5 13 3
5 14 3
9 15 7
9 16 6
12 17 5
14 18 4
*/