# 【刷题】BZOJ 1468 Tree

## Input

N（n<=40000） 接下来n-1行边描述管道，按照题目中写的输入 接下来是k

7
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10

5

## Solution

calc中如果有两点在同一子树内，那么它们在当前calc中信息是假的（它们本身之间的距离并没有那么大），所以在访问这个子树之前要先把多算的减掉

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=40000+10,inf=0x3f3f3f3f;
int n,k,e,to[MAXN<<1],nex[MAXN<<1],beg[MAXN],w[MAXN<<1],deep[MAXN],cnt,Msonsize[MAXN],size[MAXN],d[MAXN],root,finish[MAXN];
template<typename T> inline void read(T &x)
{
T data=0,w=1;
char ch=0;
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')w=-1,ch=getchar();
while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
if(x<0)putchar('-'),x=-x;
if(x>9)write(x/10);
putchar(x%10+'0');
if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void insert(int x,int y,int z)
{
to[++e]=y;
nex[e]=beg[x];
beg[x]=e;
w[e]=z;
}
inline void getroot(int x,int f,int ntotal)
{
Msonsize[x]=0;size[x]=1;
for(register int i=beg[x];i;i=nex[i])
if(to[i]==f||finish[to[i]])continue;
else
{
getroot(to[i],x,ntotal);
size[x]+=size[to[i]];
chkmax(Msonsize[x],size[to[i]]);
}
chkmax(Msonsize[x],ntotal-size[x]);
if(Msonsize[x]<Msonsize[root])root=x;
}
inline void getdeep(int x,int f)
{
deep[++cnt]=d[x];
for(register int i=beg[x];i;i=nex[i])
if(to[i]==f||finish[to[i]])continue;
else d[to[i]]=d[x]+w[i],getdeep(to[i],x);
}
inline int calc(int x,int st)
{
d[x]=st;cnt=0;
getdeep(x,0);
std::sort(deep+1,deep+cnt+1);
int l=1,r=cnt,ans=0;
while(l<r)
{
if(deep[l]+deep[r]<=k)ans+=r-l,l++;
else r--;
}
return ans;
}
inline int solve(int x)
{
int res=calc(x,0);
finish[x]=1;
for(register int i=beg[x];i;i=nex[i])
if(!finish[to[i]])
{
res-=calc(to[i],w[i]);
root=0;
getroot(to[i],x,size[to[i]]);
res+=solve(root);
}
return res;
}
int main()
{
for(register int i=1;i<n;++i)
{
int u,v,w;