Tree POJ-1741（点分治+树形DP）

题意：

给一棵树，问有多少条边的长度大于$K$。

思路：

对于一个点$x$，处理出所有其他点到它的距离，$n^2$找出经过它的所有长度加起来大于$K$的点对的数量。

但是找到的这些点对中存在不经过点$x$的点对，则减去这些点对的数量，若$v$是$x$的子节点，$w$是$v$到$x$的距离，则减去距离点$v$和大于$K-2*w$的点对的数量。

然后在树上删去点$x$。

为了降低复杂度，每次选择重心删去，保证删去后所有联通块的点的最大值最小。

代码：

//#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 2e4 + 5;
const int mod = 1e9 + 7;

struct node
{
    int v, w, next;
}edge[maxn<<1];

int n, k;
int root, minn, cnt;
int dis[maxn], siz[maxn], f[maxn];
int head[maxn],tot;
int ans;

void init()
{
    tot = 0, ans = 0;
    memset(head, -1, sizeof head);
    memset(f, 0, sizeof f);
}

void add(int u, int v, int w)
{
    edge[tot].v = v, edge[tot].w = w;
    edge[tot].next = head[u];
    head[u] = tot++;
}

int get_size(int x, int fa)//以x为根的儿子数
{
    siz[x] = 1;
    for (int i = head[x]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if (v == fa || f[v])    continue;
        siz[x] += get_size(v, x);
    }
    return siz[x];
}

void get_root(int x, int num, int fa)//找重心(最大的儿子最小)
{
    int maxx = num - siz[x];
    for (int i = head[x]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v;
        if (v == fa || f[v])    continue;
        maxx = max(maxx, siz[v]);
        get_root(v, num, x);
    }
    if (maxx < minn)
    {
        minn = maxx;
        root = x;
    }
}

void get_dis(int x, int d, int fa)//获取以x为根的距离
{
    dis[++cnt] = d;
    for (int i = head[x]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v, w = edge[i].w;
        if (v == fa || f[v])    continue;
        get_dis(v, d + w, x);
    }
}

int work(int x, int d)
{
    cnt = 0;
    get_dis(x, d, 0);
    sort(dis + 1, dis + 1 + cnt);
    int res = 0, l = 1, r = cnt;
    while (l < r)
    {
        while (l<r && dis[l] + dis[r]>k)    r--;
        res += r - l;
        l++;
    }
    return res;
}

void dfs(int x)
{
    minn = inf;
    get_size(x, 0);
    get_root(x, siz[x], 0);
    ans += work(root, 0);
    f[root] = 1;//去除这个点
    for (int i = head[root]; i != -1; i = edge[i].next)
    {
        int v = edge[i].v, w = edge[i].w;
        if (f[v])    continue;
        ans -= work(v, w);//减去不需要经过root的情况
        dfs(v);
    }
}

int main()
{
    while (~scanf("%d %d", &n, &k) && n && k)
    {
        init();
        for (int i = 1; i < n; ++i)
        {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            add(u, v, w), add(v, u, w);
        }
        dfs(1);
        printf("%d
", ans);
    }
}