POJ1741-Tree

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 24702		Accepted: 8256

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：一棵有n个点的有边权的树，问满足x到y距离小于等于m的无序点对(x,y)的个数有多少

解题思路：树分治

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cctype>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 10;
int n, m, x, y, w;
int s[maxn], nt[maxn], e[maxn], val[maxn], cnt;
int sum[maxn], mx[maxn], dis[maxn], vis[maxn], tot;

int dfs(int k, int fa, int p)
{
	int ans = 0;
	sum[k] = 1;  mx[k] = 0;
	for (int i = s[k]; ~i; i = nt[i])
	{
		if (e[i] == fa || vis[e[i]]) continue;
		int temp = dfs(e[i], k, p);
		sum[k] += sum[e[i]];
		mx[k] = max(mx[k], sum[e[i]]);
		if (mx[temp] < mx[ans]) ans = temp;
	}
	mx[k] = max(mx[k], p - sum[k]);
	return mx[k] < mx[ans] ? k : ans;
}

void dfs1(int k, int fa, int len)
{
	dis[tot++] = len;
	for (int i = s[k]; ~i; i = nt[i])
	{
		if (e[i] == fa || vis[e[i]]) continue;
		dfs1(e[i], k, len + val[i]);
	}
}

int Find(int k, int len)
{
	tot = 0;
	dfs1(k, k, len);
	sort(dis, dis + tot);
	int ans = 0;
	for (int i = 0, j = tot - 1; i < tot; i++)
	{
		while (j >= 0 && dis[i] + dis[j] > m) j--;
		if (i < j) ans += j - i;
		else break;
	}
	return ans;
}

int solve(int k, int p)
{
	int y = dfs(k, k, p), ans = Find(y, 0);
	vis[y] = 1;
	for (int i = s[y]; ~i; i = nt[i])
	{
		if (vis[e[i]]) continue;
		ans -= Find(e[i], val[i]);
		if (sum[e[i]] < sum[y]) ans += solve(e[i], sum[e[i]]);
		else ans += solve(e[i], p - sum[y]);
	}
	return ans;
}

int main()
{
	while (~scanf("%d %d", &n, &m) && (n + m))
	{
		memset(s, -1, sizeof s);
		memset(vis, 0, sizeof vis);
		cnt = 0, mx[0] = INF;
		for (int i = 1; i < n; i++)
		{
			scanf("%d%d%d", &x, &y, &w);
			nt[cnt] = s[x], s[x] = cnt, e[cnt] = y, val[cnt++] = w;
			nt[cnt] = s[y], s[y] = cnt, e[cnt] = x, val[cnt++] = w;
		}
		printf("%d\n", solve(1, n));
	}
	return 0;
}