Codeforces Round #482 (Div. 2) C Kuro and Walking Route

C. Kuro and Walking Route

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kuro is living in a country called Uberland, consisting of nn towns, numbered from 11 to nn, and n−1n−1 bidirectional roads connecting these towns. It is possible to reach each town from any other. Each road connects two towns aa and bb. Kuro loves walking and he is planning to take a walking marathon, in which he will choose a pair of towns (u,v)(u,v) (u≠vu≠v) and walk from uu using the shortest path to vv (note that (u,v)(u,v) is considered to be different from (v,u)(v,u)).

Oddly, there are 2 special towns in Uberland named Flowrisa (denoted with the index xx) and Beetopia (denoted with the index yy). Flowrisa is a town where there are many strong-scent flowers, and Beetopia is another town where many bees live. In particular, Kuro will avoid any pair of towns (u,v)(u,v) if on the path from uu to vv, he reaches Beetopia after he reached Flowrisa, since the bees will be attracted with the flower smell on Kuro’s body and sting him.

Kuro wants to know how many pair of city (u,v) he can take as his route. Since he’s not really bright, he asked you to help him with this problem.

Input

The first line contains three integers n, x and y (1≤n≤3⋅105,1≤x,y≤n1≤n≤3⋅105,1≤x,y≤n, x≠y) - the number of towns, index of the town Flowrisa and index of the town Beetopia, respectively.

n−1lines follow, each line contains two integers a and b (1≤a,b≤n1≤a,b≤n, a≠b), describes a road connecting two towns a and b.

It is guaranteed that from each town, we can reach every other town in the city using the given roads. That is, the given map of towns and roads is a tree.

Output

A single integer resembles the number of pair of towns (u,v) that Kuro can use as his walking route.

Examples

input

Copy

3 1 3
1 2
2 3

output

Copy

5

input

Copy

3 1 3
1 2
1 3

output

Copy

4

Note

On the first example, Kuro can choose these pairs:

• (1,2)(1,2): his route would be 1→2,
• (2,3)(2,3): his route would be 2→3,
• (3,2)(3,2): his route would be 3→2
• (2,1)(2,1): his route would be 2→1,
• (3,1)(3,1): his route would be 3→2→1.

Kuro can't choose pair (1,3)(1,3) since his walking route would be 1→2→31→2→3, in which Kuro visits town 11 (Flowrisa) and then visits town 33(Beetopia), which is not allowed (note that pair (3,1)(3,1) is still allowed because although Kuro visited Flowrisa and Beetopia, he did not visit them in that order).

On the second example, Kuro can choose the following pairs:

• (1,2)(1,2): his route would be 1→2,
• (2,1)(2,1): his route would be 2→1,
• (3,2)(3,2): his route would be 3→1→2,
• (3,1)(3,1): his route would be 3→1.

题意 给出一颗n个定点的树  树上有两个点 想 x,y  任意两点相互可达 但u到v的路径上先经过x再经过y是不允许的 （u，v）和（v , u）是两条不相同的路径

问满足上述条件的路径有多少条

解析

我们选定一个点作为根节点1 size[ i ]表示 以i为根节点的子树大小

若x的祖先没有y y 的祖先没有x 那么答案就是size[x]*size[y]

若x 的祖先有y  那么我们要找x的祖先 且 是y的儿子的那个节点 fa  答案就是(n-size[fa])*size[x]

同理得到另一种情况

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+ ,mod = ,inf=0x3f3f3f3f;
const double pi=acos(-1.0);
typedef long long ll;
vector<int> g[maxn];
ll siz[maxn],fa[maxn];
ll n,x,y;
int vis[maxn];
void dfs1(int k,int f)
{
    vis[k]=;
    siz[k]=;
    fa[k]=f;
    for(int i=;i<g[k].size();i++)
    {
        int u=g[k][i];
        if(vis[u]==&&u!=f)
        {
            dfs1(u,k);
            siz[k]+=siz[u];
        }
    }
}
int dfs2(int k1,int k2)
{
    if(fa[k1]==k2)
        return k1;
    else if(fa[k1]==)
        return ;
    else
        return dfs2(fa[k1],k2);
}
int main()
{
    cin>>n>>x>>y;
    for(int i=;i<n-;i++)
    {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    memset(siz,,sizeof(siz));
    memset(vis,,sizeof(vis));
    dfs1(,);
    ll ans;
    int f1=dfs2(x,y),f2=dfs2(y,x);
    //cout<<f1<<" "<<f2<<endl;
    if(f1==f2)
        ans=siz[x]*siz[y];
    else if(f1==)
        ans=(n-siz[f2])*siz[y];
    else
        ans=(n-siz[f1])*siz[x];
//    for(int i=1;i<=n;i++)
//        cout<<i<<" "<<fa[i]<<endl;
    cout<<n*(n-)-ans<<endl;
}