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- //Nguyen Huu Hoang Minh
- #include <bits/stdc++.h>
- #define sz(x) int(x.size())
- #define all(x) x.begin(),x.end()
- #define reset(x) memset(x, 0,sizeof(x))
- #define pb push_back
- #define mp make_pair
- #define fi first
- #define se second
- #define N 10005
- #define remain(x) if (x > MOD) x -= MOD
- #define ii pair<int, int>
- #define iiii pair< ii , ii >
- #define viiii vector< iiii >
- #define vi vector<int>
- #define vii vector< ii >
- #define bit(x, i) (((x) >> (i)) & 1)
- #define Task "test"
- #define int long long
- using namespace std;
- typedef long double ld;
- const int inf = 1e10;
- const int minf = -1e10;
- int t;
- int n, m;
- vector<ii> g[N];
- int d[10][N];
- int bk, p[9];
- int limits;
- void readfile()
- {
- ios_base::sync_with_stdio(false);
- cin.tie(0);cout.tie(0);
- if (fopen(Task".inp","r"))
- {
- freopen(Task".inp","r",stdin);
- //freopen(Task".out","w",stdout);
- }
- cin >> t;
- }
- void djk(int st, int di){
- d[di][st] = 0;
- priority_queue<ii, vii, greater<ii>> q;
- q.push(ii(d[di][st],st));
- while (q.size()){
- int u = q.top().se;
- int du = q.top().fi;
- q.pop();
- if (du!=d[di][u]) continue;
- for(auto i : g[u]){
- int v = i.fi;
- int uv = i.se;
- if (d[di][v] > d[di][u]+uv){
- d[di][v] = d[di][u]+uv;
- q.push(ii(d[di][v],v));
- }
- }
- }
- }
- int dp[10][300];
- void solve(){
- cin >> n >> m;
- for(int i=0; i<n; i++){
- g[i].clear();
- }
- memset(d, 0x3f3f3f3f3f3f, sizeof d);
- for(int i=1; i<=m; i++){
- int u, v, z;
- cin >> u >> v >> z;
- g[u].pb(ii(v,z));
- g[v].pb(ii(u,z));
- }
- cin >> bk;
- for(int i=1; i<=bk; i++) cin >> p[i];
- djk(0,0);
- for(int i=1; i<=bk; i++) djk(p[i],i);
- cin >> limits;
- memset(dp, -1, sizeof dp);
- int ans=0;
- for(int i=1; i<=bk; i++){
- int msk = (1<<(i-1));
- if (d[0][p[i]] + d[i][0] <= limits){
- dp[i][msk] = d[0][p[i]];
- ans = 1;
- }
- }
- int nn = (1<<bk);
- for(int msk=1; msk < nn; msk++){
- for(int i=1; i<=bk; i++){
- if (!bit(msk,i-1)) continue;
- for(int j=1; j<=bk; j++){
- if (bit(msk,j-1)) continue;
- int nmsk = msk|(1<<(j-1));
- if (dp[i][msk]!=-1 && dp[i][msk] + d[i][p[j]] + d[j][0] <= limits){
- dp[j][nmsk] = dp[i][msk] + d[i][p[j]];
- ans = max(ans,(int)__builtin_popcount(nmsk));
- }
- }
- }
- }
- cout << ans << '\n';
- }
- void proc()
- {
- while (t--){
- solve();
- }
- }
- signed main()
- {
- readfile();
- proc();
- return 0;
- }
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