Codeforces 618D Hamiltonian Spanning Tree(树的最小路径覆盖)

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
const int N=200005;
//Code begin...

struct Edge{int p, next;}edge[N<<1];
int head[N], cnt=1;
int dee[N], sum, dp[N][2];

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}
void dfs(int x, int fa){
    int siz=0, sum=0, f=-INF, s=-INF;
    for (int i=head[x]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (v==fa) continue;
        dfs(v,x); ++siz; sum+=dp[v][0];
        if (dp[v][0]-dp[v][1]>f) s=f, f=dp[v][0]-dp[v][1];
        else if (dp[v][0]-dp[v][1]>s) s=dp[v][0]-dp[v][1];
    }
    if (siz==0) dp[x][0]=dp[x][1]=1;
    else {
        if (siz==1) dp[x][0]=sum-f, dp[x][1]=sum-f;
        else dp[x][0]=sum-f-s-1, dp[x][1]=sum-f;
    }
}
int main ()
{
    int n, x, y, u, v;
    scanf("%d%d%d",&n,&x,&y);
    FO(i,1,n) scanf("%d%d",&u,&v), add_edge(u,v), add_edge(v,u), ++dee[u], ++dee[v];
    if (x>=y) {
        bool flag=false;
        FOR(i,1,n) if (dee[i]==n-1) flag=true;
        if (flag) printf("%lld\n",(LL)(n-2)*y+x);
        else printf("%lld\n",(LL)(n-1)*y);
    }
    else {
        dfs(1,0);
        printf("%lld\n",(LL)(dp[1][0]-1)*y+(LL)(n-dp[1][0])*x);
    }
    return 0;
}