解题思路：

DLX 的模板题，重复覆盖。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <assert.h>
#define FOR(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 7500;
int boys , boy[100];
struct DLX
{
    #define FF(i,A,s) for(int i = A[s];i != s;i = A[i])
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int size,col[maxn],row[maxn],s[maxn],H[100];
    bool vis[100];
	int ans[maxn],cnt;
    void init(int m)
    {
        for(int i=0;i<=m;i++)
		{
            L[i] = i - 1;R[i] = i + 1;U[i] = D[i] = i;s[i] = 0;
        }
        memset(H,-1,sizeof(H));
        L[0] = m;R[m] = 0;size = m + 1;
    }
	void link(int r,int c)
	{
         U[size] = c;D[size] = D[c];U[D[c]] = size;D[c] = size;
         if(H[r]<0)H[r] = L[size] = R[size] = size;
         else
		 {
             L[size] = H[r];R[size] = R[H[r]];
             L[R[H[r]]] = size;R[H[r]] = size;
         }
         s[c]++;col[size] = c;row[size] = r;size++;
     }
	void del(int c)
	{
        L[R[c]] = L[c] ; R[L[c]] = R[c];
        FF(i,D,c)FF(j,R,i)U[D[j]] = U[j],D[U[j]] = D[j],--s[col[j]];
    }
    void add(int c)
    {
        R[L[c]] = L[R[c]] = c;
        FF(i,U,c)FF(j,L,i)++s[col[U[D[j]] = D[U[j]] = j]];
    }
	bool dfs(int k)
	{
        if(!R[0])
		{
            cnt = k;return 1;
        }
        int c = R[0];FF(i,R,0)if(s[c] > s[i])c = i;
        del(c);
        FF(i, D, c)
        {
            FF(j, R, i) del(col[j]);
            ans[k] = row[i];if(dfs(k + 1))return true;
            FF(j,L,i) add(col[j]);
        }
        add(c);
        return 0;
    }
    void remove(int c)
    {
        FF(i, D, c)L[R[i]] = L[i],R[L[i]] = R[i];
    }
     void resume(int c)
     {
         FF(i, U, c)L[R[i]] = R[L[i]] = i;
     }
    int A()
    {
        int res = 0;
        memset(vis,0,sizeof(vis));
        FF(i, R, 0)if(!vis[i])
        {
			res++;vis[i] = 1;
			FF(j, D, i)FF(k, R, j)vis[col[k]] = 1;
		}
        return res;
    }
    void Dance(int now, int &lim)
    {
		if(now + A() > boys) return ; int tt = 0;
		for(int i=0;i<now;i++)tt += boy[ans[i]];
		if(now - tt >= lim) return;
		if(!R[0])
		{
			lim = now - tt;
			return;
		}
        int temp = INF , c;
        FF(i,R,0)if(temp >= s[i])temp = s[i] , c = i;
        FF(i, D, c)
        {
			ans[now] = row[i];
            remove(i);FF(j, R, i)remove(j);
            Dance(now + 1, lim);
            FF(j, L, i)resume(j);resume(i);
        }
    }
}dlx;
int N , D;
int d[100][100];
int main()
{
	int T , kcase = 1;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%d%d", &N, &D);dlx.init(N);
		boys = 0; FOR(i, 1, N) scanf("%d", &boy[i]) , boys += boy[i];
		FOR(i, 1, N) FOR(j, 1, N) d[i][j] = (i == j ? 0 : INF);
		FOR(i, 1, N-1)
		{
			int u , v , w;
			scanf("%d%d%d",&u,&v,&w);
			d[u][v] = d[v][u] = w;
		}
		
		FOR(k, 1, N)
		FOR(i, 1, N)
		FOR(j, 1, N)
			if(d[i][j] > d[i][k] + d[k][j]) 
				d[i][j] = d[i][k] + d[k][j];
		
		FOR(i, 1, N) FOR(j, 1, N)
		if(d[i][j] <= D) dlx.link(i, j);
		int ans = INF;
		dlx.Dance(0, ans);
		if(ans > boys) ans = -1;
		printf("Case #%d: %d\n", kcase++, ans);
	}
	return 0;
}

HDU 4735 Little Wish~ lyrical step~（DLX , 重复覆盖）