Kpath

Suppose we want to count how many ancestor-descendant paths have a cost of at most x (arbitrarily fixed). In order to do this, we can compute:

• ancestor[j][i] = the 2^jth ancestor of vertex i (or 0 if i has less than 2^j ancestors)
• cost[j][i] = the cost of the path from i to ancestor[j][i]

Having these two tables, we can climb logarithmically from any vertex to its highest ancestor such that the cost of the path between them does not exceed x. Therefore we can find the answer for a fixed x in O(N * log(N)).

Now, we need to find the lowest x such that there exist at least K paths with a cost of at most x. We can apply a binary search and solve the problem in O(N * log(N) * log(N * MAX_WEIGHT)).

Further reading: Stramosi (infoarena).

Sergiu Pușcaș - C++11

    1 #include <bits/stdc++.h>
    2 using namespace std;
    3 
    4 typedef long long ll;
    5 const int nmax = 100005;
    6 const int logmax = 18;
    7 
    8 ll k;
    9 int n, a, b, x;
   10 int cost[logmax][nmax], ancestor[logmax][nmax];
   11 bool seen[nmax];
   12 vector <pair <int, int>> v[nmax];
   13 
   14 void explore(int x) {
   15     seen[x] = true;
   16 
   17     for(auto t:v[x])
   18         if(!seen[t.first]) {
   19             cost[0][t.first] = t.second;
   20             ancestor[0][t.first] = x;
   21             explore(t.first);
   22         }
   23 }
   24 
   25 int climb(int i, ll energy) {
   26     if(ancestor[0][i] == 0 || cost[0][i] > energy)
   27         return 0;
   28 
   29     int j=0;
   30     while(ancestor[j+1][i] != 0 && cost[j+1][i] <= energy)
   31         j++;
   32 
   33     return (1<<j) + climb(ancestor[j][i], energy - cost[j][i]);
   34 }
   35 
   36 ll how_many_paths(ll energy) {
   37     ll ret = 0;
   38     for(int i=2; i<=n; i++)
   39         ret += climb(i, energy);
   40 
   41     return ret;
   42 }
   43 
   44 int main() {
   45     cin>>n>>k;
   46 
   47     for(int i=1; i<n; i++) {
   48         cin>>a>>b>>x;
   49 
   50         v[a].push_back(make_pair(b, x));
   51         v[b].push_back(make_pair(a, x));
   52     }
   53 
   54     explore(1);
   55 
   56     for(int j=1; (1<<j)<=n; j++)
   57         for(int i=1; i<=n; i++) {
   58             ancestor[j][i] = ancestor[j-1][ancestor[j-1][i]];
   59             cost[j][i] = cost[j-1][i] + cost[j-1][ancestor[j-1][i]];
   60         }
   61 
   62     ll lo = 0, hi = 1000000007LL, mid;
   63     while(hi - lo > 1) {
   64         mid = lo + (hi - lo) / 2;
   65 
   66         if(how_many_paths(mid) >= k)
   67             hi = mid;
   68         else
   69             lo = mid;
   70     }
   71 
   72     if(how_many_paths(hi) < k)
   73         hi = -1;
   74 
   75     cout<<hi<<"\n";
   76     return 0;
   77 }