description: <image>
  
  William really likes puzzle kits. For one of his birthdays, his friends gifted him a complete undirected edge-weighted graph consisting of n vertices.
  
  He wants to build a spanning tree of this graph, such that for the first k vertices the following condition is satisfied: the degree of a vertex with index i does not exceed d_i. Vertices from k + 1 to n may have any degree.
  
  William wants you to find the minimum weight of a spanning tree that satisfies all the conditions.
  
  A spanning tree is a subset of edges of a graph that forms a tree on all n vertices of the graph. The weight of a spanning tree is defined as the sum of weights of all the edges included in a spanning tree.
  
  Input
  
  The first line of input contains two integers n, k (2 ≤ n ≤ 50, 1 ≤ k ≤ min(n - 1, 5)).
  
  The second line contains k integers d_1, d_2, …, d_k (1 ≤ d_i ≤ n). 
  
  The i-th of the next n - 1 lines contains n - i integers w_{i,i+1}, w_{i,i+2}, …, w_{i,n} (1 ≤ w_{i,j} ≤ 100): weights of edges (i,i+1),(i,i+2),…,(i,n).
  
  Output
  
  Print one integer: the minimum weight of a spanning tree under given degree constraints for the first k vertices. 
  
  Example
  
  Input
  
  
  10 5
  5 3 4 2 1
  29 49 33 12 55 15 32 62 37
  61 26 15 58 15 22 8 58
  37 16 9 39 20 14 58
  10 15 40 3 19 55
  53 13 37 44 52
  23 59 58 4
  69 80 29
  89 28
  48
  
  
  Output
  
  
  95
solution: #include <bits/stdc++.h>
  using namespace std;
  int nw[52], d[52], sz[52][52], fa[52];
  int getv(int x) { return x == fa[x] ? x : fa[x] = getv(fa[x]); }
  bool merge(int x, int y) {
    x = getv(x);
    y = getv(y);
    if (x == y) return false;
    fa[x] = y;
    return true;
  }
  int random(int n) {
    int he = 0;
    for (int i = 1; i <= n; i++) nw[i] = 0;
    vector<pair<int, int> > ve;
    for (int i = 1; i < n; i++) {
      he += sz[i][n], nw[i] = 1;
      ve.push_back(make_pair(i, n));
    }
    int ans = he;
    nw[n] = n - 1;
    double T = 100 * n;
    while (T > 1e-5) {
      int od = he, i = rand() % (n - 1);
      int oa = ve[i].first, ob = ve[i].second;
      nw[oa] -= 1;
      nw[ob] -= 1;
      he -= sz[oa][ob];
      ve.erase(ve.begin() + i);
      for (int i = 1; i <= n; i++) fa[i] = i;
      for (int i = 0; i < n - 2; i++) merge(ve[i].first, ve[i].second);
      int a, b;
      while (1) {
        a = rand() % (n - 1) + 1;
        b = rand() % (n - a) + a + 1;
        if (nw[a] < d[a] && nw[b] < d[b] && merge(a, b)) {
          he += sz[a][b];
          ve.push_back(make_pair(a, b));
          nw[a] += 1;
          nw[b] += 1;
          break;
        }
      }
      if (he < ans) ans = he;
      if (he > od && rand() % 10000 >= exp((od - he) / T) * 1e4) {
        he = od;
        ve.pop_back();
        ve.push_back(make_pair(oa, ob));
        nw[a] -= 1;
        nw[b] -= 1;
        nw[oa] += 1;
        nw[ob] += 1;
      }
      T *= 0.999995;
    }
    return ans;
  }
  int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= k; i++) scanf("%d", &d[i]);
    for (int i = k + 1; i <= n; i++) d[i] = 1e9;
    for (int i = 1; i < n; i++) {
      for (int j = i + 1; j <= n; j++) {
        scanf("%d", &sz[i][j]);
        sz[j][i] = sz[i][j];
      }
    }
    int ans = random(n);
    printf("%d", ans);
    return 0;
  }
  