Poj-1741 Tree(树的点分治)

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.

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.

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

LouTiancheng@POJ

题意：给定一颗带边权的树，问数上距离不超过k的不同点对有多少个。

分析：树的点分治，所有树上的路径要么经过根节点要么属于一颗子树,我们处理出经过根节点的值，然后对所有子树递归处理，每次选择重心作为根的话最多走log(n)层，每层log(n)*n的排序复杂度，总复杂度n*log(n)*log(n)。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MOD 1000000007
#define MAXN 10005
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
vector<pii> G[MAXN];
vector<int> b;
int n,k,ans,root,size[MAXN];
bool jud[MAXN];
int dfs(int u,int fa,int tot)
{
bool flag = true;
size[u] = 1;
for(int i = 0;i < G[u].size();i++)
{
pii vv = G[u][i];
if(vv.first != fa && !jud[vv.first])
{
int v = vv.first;
int tmp = dfs(v,u,tot);
if(tmp) return tmp;
size[u] += size[v];
if(size[v] > tot/2 ) flag = false;
}
}
if(tot - size[u] > tot/2) flag = false;
return flag ? u : 0;
}
int got_size(int u,int fa)
{
int num = 1;
for(int i = 0;i < G[u].size();i++)
{
pii vv = G[u][i];
if(vv.first != fa && !jud[vv.first]) num += got_size(vv.first,u);
}
return num;
}
void dfs2(int u,int fa,int deep)
{
b.push_back(deep);
for(int i = 0;i < G[u].size();i++)
{
pii vv = G[u][i];
if(vv.first != fa && !jud[vv.first]) dfs2(vv.first,u,deep+vv.second);
}
}
void deal(int u,int deep,int x)
{
b.clear();
dfs2(u,-1,deep);
sort(b.begin(),b.end());
int tot = 0,sum = b.size(),r = -1;
for(int i = sum-1;i;i--)
{
while(b[i] + b[r+1] <= k && r+1 < i) r++;
if(i == r) r--;
tot += r+1;
}
ans += tot*x;
}
void calc(int u)
{
deal(u,0,1);
jud[u] = true;
for(int i = 0;i < G[u].size();i++)
{
pii vv = G[u][i];
if(!jud[vv.first])
{
int v = vv.first;
deal(v,vv.second,-1);
int root = dfs(v,u,got_size(v,u));
calc(root);
}
}
}
int main()
{
while(scanf("%d%d",&n,&k) && n+k)
{
ans = 0;
memset(jud,0,sizeof(jud));
for(int i = 1;i <= n;i++) G[i].clear();
for(int i = 1;i < n;i++)
{
int u,v,val;
scanf("%d%d%d",&u,&v,&val);
G[u].push_back(make_pair(v,val));
G[v].push_back(make_pair(u,val));
}
root = dfs(1,0,n);
calc(root);
cout<<ans<<endl;
}
}