Submission #1059610

#	Time	Username	Problem	Language	Result	Execution time	Memory
1059610		vjudge1	Cities (BOI16_cities)	C++17	74 / 100	6054 ms	172684 KiB

cities

#include <bits/stdc++.h>

using namespace std;

//#define int long long
#define ll long long
#define ii pair<int, int>
#define tp tuple<ll, int, int>
#define fi first
#define se second
#define all(x) x.begin(), x.end()

const int N = 2e5 + 5,
          mod = 1e9 + 7;
const ll INF = 2e16;

// code

int n, k, m, a[6];
ll dp[N][33];
vector<ii> ad[N];
vector<int> nxt[33];
priority_queue<tp, vector<tp>, greater<>> pq;

int id(int i){
  for(int j = 0; j < k; j++) if(a[j] == i) return j;
  return -1;
}

bool check(int msk1, int msk2){
  for(int i = 0; i < (1 << k); i++) if((msk1 >> i & 1) && (msk2 >> i & 1)) return false;
  return true;
}

int32_t main(){
  cin.tie(0)->sync_with_stdio(0);
//  #define task "test"
//  if(fopen(task ".inp", "r")){
//    freopen(task ".inp", "r", stdin);
//    freopen(task ".out", "w", stdout);
//  }

  cin >> n >> k >> m;

  for(int i = 0; i < k; i++) cin >> a[i];
  for(int i = 0; i < m; i++){
    int a, b, c; cin >> a >> b >> c;
    ad[a].push_back({b, c});
    ad[b].push_back({a, c});
  }
  for(int msk = 0; msk < (1 << k); msk++)
    for(int _msk = 0; _msk < (1 << k); _msk++)
      if(check(msk, _msk)) nxt[msk].push_back(_msk);

  for(int i = 1; i <= n; i++)
    for(int j = 1; j < (1 << k); j++) dp[i][j] = INF;

  for(int i = 0; i < k; i++){
    dp[a[i]][1 << i] = 0;
    pq.push({0, 1 << i, a[i]});
  }

  while(pq.size()){
    auto [d, msk, u] = pq.top(); pq.pop();

    if(d != dp[u][msk]) continue;

//    cerr << u << " " << d << " " << msk << endl;

    for(auto [v, c] : ad[u]){
      int bit = id(v);

      if(bit != -1 && !(msk >> bit & 1))
        if(d + c < dp[v][msk|(1 << bit)])
          pq.push({dp[v][msk|(1 << bit)] = d + c, msk|(1 << bit), v});

      for(auto _msk : nxt[msk])
        if(d + c + dp[v][_msk] < dp[v][msk|_msk])
          pq.push({dp[v][msk|_msk] = d + c + dp[v][_msk], msk|_msk, v});
    }
  }

  cout << dp[a[0]][(1 << k)-1];

  return 0;
}

• Subtask 1: 22 / 22
• Subtask 2: 14 / 14
• Subtask 3: 15 / 15
• Subtask 4: 23 / 23
• Subtask 5: 0 / 26