time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
A group of n cities is connected by a network of roads. There is an undirected road between every pair of cities, so there are roads in total. It takes exactly y seconds to traverse any single road.
A spanning tree is a set of roads containing exactly n?-?1 roads such that it‘s possible to travel between any two cities using only these roads.
Some spanning tree of the initial network was chosen. For every road in this tree the time one needs to traverse this road was changed from y to x seconds. Note that it‘s not guaranteed that x is smaller than y.
You would like to travel through all the cities using the shortest path possible. Given n, x, y and a description of the spanning tree that was chosen, find the cost of the shortest path that starts in any city, ends in any city and visits all cities exactly once.
Input
The first line of the input contains three integers n, x and y (2?≤?n?≤?200?000,?1?≤?x,?y?≤?109).
Each of the next n?-?1 lines contains a description of a road in the spanning tree. The i-th of these lines contains two integers ui and vi (1?≤?ui,?vi?≤?n) — indices of the cities connected by the i-th road. It is guaranteed that these roads form a spanning tree.
Output
Print a single integer — the minimum number of seconds one needs to spend in order to visit all the cities exactly once.
Examples
Input
Copy
5 2 3
1 2
1 3
3 4
5 3
Output
Copy
9
Input
Copy
5 3 2
1 2
1 3
3 4
5 3
Output
Copy
8
Note
In the first sample, roads of the spanning tree have cost 2, while other roads have cost 3. One example of an optimal path is .

In the second sample, we have the same spanning tree, but roads in the spanning tree cost 3, while other roads cost 2. One example of an optimal path is .

題解:給了一個無向完全圖,還有一個生成樹,在生成樹上的邊的權值為x,不在生成樹上的權值為y,讓你求一條最短路徑(閉環的),經過所有的點,有且僅有一次經過.

根據x,y的大小先分類討論:

1.x>y時,,盡量走不在樹上的邊.在普通情況下,總能找到不在上的邊,所以cost就為y*(n-1).

特殊情況,如:菊花圖

po主親自繪制.

這個特判即可,詳見代碼.

2.x<y時,盡量走在樹上的邊,dfs跑一下,記錄一下樹上可用的邊,找出最長的那一條,剩下的用y來連接,cos為cntx+(n-1-cnt)y.

#include <bits/stdc++.h>
const int N=2e5+5;
typedef long long ll;
using namespace std;
vector<int> v[N];
int du[N];
ll cnt=0;
bool dfs(int n,int m){
    int left=2;
    for(unsigned i=0;i<v[n].size();i++){
        int w=v[n][i];
        if(w==m) continue;
        bool flag=dfs(w,n);
        if(flag&&left>0){
            cnt++;
            left--;
        }
    }
    return left>0;
}
int main()
{
    ll n,x,y;
    scanf("%I64d%I64d%I64d",&n,&x,&y);
    for(int i=1;i<n;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        du[a]++;
        du[b]++;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    ll ans=0;
    if(x<=y){
        dfs(1,-1);
        ans=cnt*x+(n-1-cnt)*y;
    }
    else{
        int flag=0;
        for(int i=1;i<=n;i++){
            if(du[i]==n-1) flag=1;
        }
        if(flag) ans=(n-2)*y+x; else ans=(n-1)*y;

    }
    printf("%I64d\n",ans);
    //cout << "Hello world!" << endl;
    return 0;
}
 

Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) D. Hamiltonian Spanning Tree