BZOJ1468:Tree(点分治)

Description

给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

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

Output

一行，有多少对点之间的距离小于等于k

Sample Input

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

Sample Output

5

Solution

点分治模板

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define N (40000+10)
 using namespace std;
 struct node
 {
     int to,next,len;
 }edge[N*];
 int n,k,sum,root,ans;
 int head[N],num_edge;
 int depth[N],d[N],size[N],maxn[N];
 bool vis[N];
 void add(int u,int v,int l)
 {
     edge[++num_edge].to=v;
     edge[num_edge].len=l;
     edge[num_edge].next=head[u];
     head[u]=num_edge;
 }

 void Get_root(int x,int fa)
 {
     size[x]=; maxn[x]=;
     for (int i=head[x];i!=;i=edge[i].next)
         if (edge[i].to!=fa && !vis[edge[i].to])
         {
             Get_root(edge[i].to,x);
             size[x]+=size[edge[i].to];
             maxn[x]=max(maxn[x],size[edge[i].to]);
         }
     maxn[x]=max(maxn[x],sum-size[x]);
     if (maxn[x]<maxn[root]) root=x;
 }

 void Get_depth(int x,int fa)
 {
     depth[++depth[]]=d[x];
     for (int i=head[x];i!=;i=edge[i].next)
         if (edge[i].to!=fa && !vis[edge[i].to])
         {
             d[edge[i].to]=d[x]+edge[i].len;
             Get_depth(edge[i].to,x);
         }
 }

 int Calc(int x,int cost)
 {
     d[x]=cost; depth[]=;
     Get_depth(x,);
     sort(depth+,depth+depth[]+);
     int l=,r=depth[],cnt=;
     while (l<r)
         if (depth[l]+depth[r]<=k)
             cnt+=r-l,l++;
         else
             r--;
     return cnt;
 }

 void Solve(int x)
 {
     ans+=Calc(x,);
     vis[x]=true;
     for (int i=head[x];i!=;i=edge[i].next)
         if (!vis[edge[i].to])
         {
             ans-=Calc(edge[i].to,edge[i].len);
             sum=size[edge[i].to];
             root=;
             Get_root(edge[i].to,);
             Solve(root);
         }
 }

 int main()
 {
     int u,v,l;
     scanf("%d",&n);
     sum=maxn[]=n;
     for (int i=;i<=n-;++i)
     {
         scanf("%d%d%d",&u,&v,&l);
         add(u,v,l); add(v,u,l);
     }
     scanf("%d",&k);
     Get_root(,);
     Solve(root);
     printf("%d",ans);
 }