题目链接：点击打开链接

题意：

给定n个点的有根树（0为根），

下面给出边和边权

一个整数q表示q个询问

每个询问一个数字x ，表示有一个人从根开始走，行走距离不超过x且使得走过不相同的点最多。

问最多能走多少个点。

思路:

dp[i][j][0]表示以i为根的子树，以i为起点走了j个不同点且回到i的最小花费。

dp[i][j][1]表示不需要回到i的最小花费。

转移的时候就是一个背包

import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Queue;

public class Main {
	int n, q;
	int[] siz = new int[N];
	int[][][] dp = new int[N][N][2];//dp[i][j][0]表示以i为子树 走了j个点后回到i的最小花费，1表示不回到i
	int[][] tmp = new int[N][2];
	void input(){
		init_edge();
		for(int i = 1, u, v, dis; i < n; i++){
			u = cin.nextInt(); v = cin.nextInt(); dis = cin.nextInt();
			add(u, v, dis);
			add(v, u, dis);
		}
		q = cin.nextInt();
	}
	void dfs(int u, int fa){
		dp[u][1][0] = dp[u][1][1] = 0;
		siz[u] = 1;
		
		for(int i = head[u]; i != -1; i = edge[i].nex){
			int v = edge[i].to; if(v == fa)continue;
			dfs(v, u);
			siz[u] += siz[v];
			
			for(int j = siz[u]; j > 1; j--){
				for(int k = 1; k <= siz[v] && j-k>=0; k++)
				{
					dp[u][j][0] = min(dp[u][j][0], dp[u][j-k][0]+dp[v][k][0]+edge[i].dis*2);
					dp[u][j][1] = min(dp[u][j][1], dp[u][j-k][0]+dp[v][k][1]+edge[i].dis);
					dp[u][j][1] = min(dp[u][j][1], dp[u][j-k][1]+dp[v][k][0]+edge[i].dis*2);					
				}			
			}
		}
//		out.print(u+" back:");	for(int i = 1; i <= siz[u]; i++)out.print(dp[u][i][0]+ " "); out.println();out.print(u+" not back:");	for(int i = 1; i <= siz[u]; i++)out.print(dp[u][i][1]+ " "); out.println();
	}
	void work() {
		int Cas = 1;
		while(true){
			n = cin.nextInt(); if(n == 0)break;
			input();
			for(int i = 0; i <= n; i++) for(int j = 0; j <= n; j++)dp[i][j][0] = dp[i][j][1] = inf;
			dfs(0, 0);
			out.println("Case "+(Cas++)+":");
			while(q-->0){
				int x = cin.nextInt(), ans = 1;
				for(int i = 2; i <= n; i++)
					if(x>=dp[0][i][1])
						ans = i;
				out.println(ans);
			}
		}
	}

	Main() {
		cin = new Scanner(System.in);
		out = new PrintWriter(System.out);
	}

	public static void main(String[] args) {
		Main e = new Main();
		e.work();
		out.close(); 
	}

	public Scanner cin;
	public static PrintWriter out;
	static int N = 505;
	static int M = 505;
	DecimalFormat df=new DecimalFormat("0.0000");
	static int inf = (int) 1e9 + 7;
	static long inf64 = (long) 1e18;
	static double eps = 1e-8;
	static int mod = 1000000007 ;

	class Edge{
		int from, to, dis, nex;
		Edge(){}
		Edge(int from, int to, int dis, int nex){
			this.from = from;
			this.to = to;
			this.dis = dis;
			this.nex = nex;
		}
	}
	Edge[] edge = new Edge[M<<1];
	int[] head = new int[N];
	int edgenum;
	void init_edge(){for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}
	void add(int u, int v, int dis){
		edge[edgenum] = new Edge(u, v, dis, head[u]);
		head[u] = edgenum++;
	}/*
	int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
		int pos = r;
		r--;
		while (l <= r) {
			int mid = (l + r) >> 1;
			if (A[mid] <= val) {
				l = mid + 1;
			} else {
				pos = mid;
				r = mid - 1;
			}
		}
		return pos;
	}/**/

	int Pow(int x, int y) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	double Pow(double x, int y) {
		double ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	int Pow_Mod(int x, int y, int mod) {
		int ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}

	long Pow(long x, long y) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			y >>= 1;
			x = x * x;
		}
		return ans;
	}
	

	long Pow_Mod(long x, long y, long mod) {
		long ans = 1;
		while (y > 0) {
			if ((y & 1) > 0)
				ans *= x;
			ans %= mod;
			y >>= 1;
			x = x * x;
			x %= mod;
		}
		return ans;
	}

	int gcd(int x, int y){
		if(x>y){int tmp = x; x = y; y = tmp;}
		while(x>0){
			y %= x;
			int tmp = x; x = y; y = tmp;
		}
		return y;
	}
	int max(int x, int y) {
		return x > y ? x : y;
	}

	int min(int x, int y) {
		return x < y ? x : y;
	}

	double max(double x, double y) {
		return x > y ? x : y;
	}

	double min(double x, double y) {
		return x < y ? x : y;
	}

	long max(long x, long y) {
		return x > y ? x : y;
	}

	long min(long x, long y) {
		return x < y ? x : y;
	}

	int abs(int x) {
		return x > 0 ? x : -x;
	}

	double abs(double x) {
		return x > 0 ? x : -x;
	}

	long abs(long x) {
		return x > 0 ? x : -x;
	}

	boolean zero(double x) {
		return abs(x) < eps;
	}
}

UVALive 4015 Caves 树形背包