Codeforces 618D Hamiltonian Spanning Tree (DFS)

Topic Link: "Codeforces 618D"

A spanning tree consisting of n nodes and (n-1) edges is given, and the spanning tree is made from the complete graph containing the n nodes, the weights on the top of the spanning tree are X, the weights of the edges in the full graph but not on the spanning tree are Y, and the shortest path to iterate through all the points

Each point in the shortest path has a maximum of two edges, divided into two types of discussion:

1. X<y ==> makes the edges in the shortest path as many as possible in the spanning tree

Define a point as the starting point, Dfs runs one side, the edge of the running edge record node V There are still several sides can be connected (that is, left in the code), and ANS to record the number of edges on the spanning tree

2. X>=y ==> make the edges in the shortest path as many as possible in the spanning tree

When the number of edges on the spanning tree is n-1, there must be 1 edges on the spanning tree, or 0.

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <
String> #include <vector> using namespace std;
#define LL __int64 int n, x, y;
int ans;
vector<int>vec[200020];
    int dfs (int u, int p) {int left=2;
    	for (int i=0; i<vec[u].size (); i++) {int v = vec[u][i];
    	if (v==p) continue;
    	int x = DFS (v, u);
    		if (left>0 && x==1) {ans++;
    	left--; }} return left>0?
1:0;
	} int main () {cin>>n>>x>>y;
	int u, v;
	ll S1, S2;
		for (int i=0; i<n-1; i++) {cin>>u>>v;
		Vec[u].push_back (v);
	Vec[v].push_back (U);
	} ans=0;
		if (x<y) {DFS (1,-1);
		S1 = (ll) Ans*x + (LL) (n-ans-1) *y;
	cout<<s1<<endl;
				} else {for (int i=1; i<=n; i++) {if ((int) vec[i].size () ==n-1) {Ans=1;
			Break
		}} s2= (LL) Ans*x + (LL) (n-ans-1) *y;
	cout<<s2<<endl; 
} return 0; }