CPP Program Edges in MST – Advanced Graphs

  • You are given a connected weighted undirected graph without any loops and multiple edges.
  • Let us remind you that a graph’s spanning tree is defined as an acyclic connected subgraph of the given graph that includes all of the graph’s vertexes. The weight of a tree is defined as the sum of the weights of the edges that the given tree contains.
  • The minimum spanning tree (MST) of a graph is defined as the graph’s spanning tree having the minimum possible weight. For any connected graph obviously exists the minimum spanning tree, but in the general case, a graph’s minimum spanning tree is not unique.
  • Your task is to determine the following for each edge of the given graph: whether it is either included in any MST, included at least in one MST, or not included in any MST.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105) – the number of the graph’s vertexes and edges, correspondingly. Then follow m lines, each of them contains three integers – the description of the graph’s edges as “ai bi wi” (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106, ai ≠ bi), where ai and bi are the numbers of vertexes connected by the i-th edge, wi is the edge’s weight. It is guaranteed that the graph is connected and doesn’t contain loops or multiple edges.

Output

Print m lines — the answers for all edges. If the i-th edge is included in any MST, print “any”; if the i-th edge is included at least in one MST, print “at least one”; if the i-th edge isn’t included in any MST, print “none”. Print the answers for the edges in the order in which the edges are specified in the input.

Sample Input 1:

4 5
1 2 101
1 3 100
2 3 2
2 4 2
3 4 1

Sample Output 1:

none
any
at least one
at least one
any

Sample Input 2:

3 3
1 2 1
2 3 1
1 3 2

Sample Output 2:

any
any
none

Sample Input 3:

3 3
1 2 1
2 3 1
1 3 1

Sample Output 3:

at least one at least one at least one

Note:

  • In the second sample, the MST is unique for the given graph: it contains two first edges.
  • In the third sample, any two edges form the MST for the given graph. That means that each edge is included at least in one MST.

Code:

#include <iostream> #include <vector> #include <algorithm> #define N 100001 using namespace std; int n, m, x[N], y[N], z[N], p[N]; int ans[N], f[N], h[N], pe[N], d[N]; vector<pair<int, int>> v[N]; bool cmp(const int &x, const int &y) { return z[x] < z[y]; } int par(int x) { while (pe[x]) x = pe[x]; return x; } void uni(int x, int y) { x = par(x); y = par(y); v[x].clear(); v[y].clear(); f[x] = 0; f[y] = 0; if (x == y) return; if (h[x] > h[y]) pe[y] = x; else { pe[x] = y; if (h[x] == h[y]) h[y]++; } } void add_edge(int x, int y, int i) { if (x == y) return; ans[i] = 1; v[x].push_back({y, i}); v[y].push_back({x, i}); } void kruskal(int c, int g, int h) { f[c] = true; d[c] = h; for (pair<int, int> i : v[c]) if (!f[i.first]) { kruskal(i.first, i.second, h + 1); d[c] = min(d[c], d[i.first]); } else if (i.second != g) d[c] = min(d[c], d[i.first]); if (d[c] == h) ans[g] = 2; } int main() { cin >> n >> m; for (int i = 1; i <= m; i++) { cin >> x[i] >> y[i] >> z[i]; p[i] = i; } sort(p + 1, p + m + 1, cmp); for (int i = 1; i <= m;) { int j; for (j = i; z[p[j]] == z[p[i]]; j++) add_edge(par(x[p[j]]), par(y[p[j]]), p[j]); for (j = i; z[p[j]] == z[p[i]]; j++) { int k = par(x[p[j]]); if (!f[k]) kruskal(k, 0, 0); } for (j = i; z[p[j]] == z[p[i]]; j++) uni(x[p[j]], y[p[j]]); i = j; } for (int i = 1; i <= m; i++) if (!ans[i]) cout << "none" << endl; else if (ans[i] == 1) cout << "at least one" << endl; else cout << "any" << endl; return 0; }
Code language: C++ (cpp)

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