poj 1741 树的点分治(入门)

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18205		Accepted: 5951

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

/*
poj 1741 树的点分治(入门)

problem:
给一棵边带权树，问两点之间的距离小于等于K的点对有多少个

solve:
随便找的博客学习下
大致思路,对当前树先求其重心为根节点,然后找出所有点到根节点的距离. 然后就能计算出有多少的的和
<= k. 但是这两个点有可能来自同一个子树,所以在后面计算中减去,就能计算出过当前根节点的所有对.
以同样的方法计算子树的情况,递推下去就行.
重心是为了让最大子树的节点数最小,从而增加效率.
(个人理解)
hhh-2016-08-22 20:00:18
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
#define inf 0x3FFFFFFFFFFFFFFFLL
using namespace std;
const int maxn = 100010;

struct node
{
int to,w;
node() {};
node(int v,int _w):to(v),w(_w) {};
};

vector<node> q[maxn];
int n,k,s[maxn],f[maxn],root,d[maxn],ans,limit;
int Size;
bool vis[maxn];
vector<int> ta;

void get_root(int now,int fa)
{
int v;
s[now] = 1,f[now] = 0;
for(int i = 0;i < q[now].size();i++)
{
if( (v=q[now][i].to) == fa || vis[v])
continue;
get_root(v,now);
s[now] += s[v];
f[now] = max(f[now],s[v]);
}
f[now] = max(f[now],Size - s[now]);
if(f[now] < f[root]) root = now;
}

void dfs(int now,int fa)
{
int v;
ta.push_back(d[now]);
s[now] = 1;
for(int i = 0;i < q[now].size();i++)
{
if( (v=q[now][i].to) == fa || vis[v])
continue;
d[v] = d[now] + q[now][i].w;
dfs(v,now);
s[now] += s[v];
}
}

int cal(int now,int begi)
{
ta.clear(),d[now] = begi;
dfs(now,0);
sort(ta.begin(),ta.end());
int cnt = 0;
for(int l = 0,r=ta.size()-1;l<r;)
{
if(ta[l] + ta[r] <= limit) cnt += (r-l++);
else r --;
}
return cnt;
}

void work(int now)
{
int v;
ans += cal(now,0);
vis[now] = 1;
for(int i = 0;i < q[now].size(); i++)
{
if(!vis[v = q[now][i].to])
{
ans -= cal(v,q[now][i].w);
f[0] = Size = s[v];
get_root(v,root = 0);
work(root);
}
}
}

int main()
{
while(scanf("%d%d",&n,&limit) == 2)
{
if(!n && !limit)
break;
for(int i = 0;i <= n;i++) q[i].clear();
memset(vis,0,sizeof(vis));

int a,b,c;
for(int i = 1;i < n;i++)
{
scanf("%d%d%d",&a,&b,&c);
q[a].push_back(node(b,c));
q[b].push_back(node(a,c));
}
f[0] = Size = n;
get_root(1,root = 0);
ans = 0;
work(root);
printf("%d\n",ans);
}
return 0;
}