传送门

题意

给你一棵树，有 n n n个点，边权都为 1 1 1，问你树上有多少对点的距离为 k k k

分析

两种思路，第一种思路是树形DP，为 f [ u ] [ x ] f[u][x] f[u][x]维护树上到点 u u u距离为 x x x的点的数量，向下跑一遍，然后向上维护一下就好了
还有一种点分治的做法下次补上

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 50000 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n,k;
ll f[N][510];
int h[N],e[M],ne[M],idx;

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u,int fa){
    f[u][0] = 1;
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa) continue;
        dfs(j,u);
        for(int x = 1;x <= k;x++) f[u][x] += f[j][x - 1];
    }
}

void dfs_(int u,int fa){
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == fa) continue;
        for(int x = k;x;x--) {
            f[j][x] += f[u][x - 1];
            if(x > 1) f[j][x] -= f[j][x - 2];
        }
        dfs_(j,u);
    }
}

int main() {
    memset(h,-1,sizeof h);
    read(n),read(k);
    for(int i = 1;i < n;i++){
        int x,y;
        read(x),read(y);
        add(x,y),add(y,x);
    }
    dfs(1,0);
    dfs_(1,0);
    ll ans = 0;
    for(int i = 1;i <= n;i++) ans += f[i][k];
    dl(ans / 2ll);
    return 0;
}