解题思路：
DLX 的模板题。反复覆盖。
#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;
}