Kuro and Walking Route

Posted 2020-11-06 行远山

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

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

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

Input

The first line contains three integers $\mathrm{nn, xx and yy (1\le n\le 3?105,1\le x,y\le n1\le n\le 3?105,1\le x,y\le n, x\ne yx\ne y)\; -\; the\; number\; of\; towns,\; index\; of\; the\; town\; Flowrisa\; and\; index\; of\; the\; town\; Beetopia,\; respectively.}$

$\mathrm{n?1n?1 lines\; follow,\; each\; line\; contains\; two\; integers aa and bb (1\le a,b\le n1\le a,b\le n, a\ne ba\ne b),\; describes\; a\; road\; connecting\; two\; towns aa and bb.}$

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 $(u,v)(u,v) that\; Kuro\; can\; use\; as\; his\; walking\; route.$

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:

• $(1,2)(1,2):\; his\; route\; would\; be 1\to 21\to 2,$
• $(2,3)(2,3):\; his\; route\; would\; be 2\to 32\to 3,$
• $(3,2)(3,2):\; his\; route\; would\; be 3\to 23\to 2,$
• $(2,1)(2,1):\; his\; route\; would\; be 2\to 12\to 1,$
• $(3,1)(3,1):\; his\; route\; would\; be 3\to 2\to 13\to 2\to 1.$

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

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

• $(1,2)(1,2):\; his\; route\; would\; be 1\to 21\to 2,$
• $(2,1)(2,1):\; his\; route\; would\; be 2\to 12\to 1,$
• $(3,2)(3,2):\; his\; route\; would\; be 3\to 1\to 23\to 1\to 2,$
• $(3,1)(3,1):\; his\; route\; would\; be 3\to 13\to 1.$

答案：以x为根，搜y有多少个子节点，再以y为根，搜x有多少个子节点。总方案数减去两个子节点乘积，就是答案了。

#include <iostream>
#include <bits/stdc++.h>
#define maxn 600005
typedef long long ll;
using namespace std;
struct edge
{
    int u;
    int v;
    int next;
}edge[maxn];
int head[maxn];
int cnt=0;
int x,y;
void add(int u,int v)
{
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int sizes[maxn];
bool vis[maxn];
void dfs(int num)
{    int flag=0;
     for(int i=head[num];i!=-1;i=edge[i].next)
     {
       if(!vis[edge[i].v])
       {   vis[edge[i].v]=true;
           dfs(edge[i].v);
           sizes[num]+=sizes[edge[i].v];
           flag=1;
       }
     }
     if(flag==0)
       {
           sizes[num]=1;
       }
    else
    {
        sizes[num]++;
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    ll n;
    scanf("%I64d%d%d",&n,&x,&y);
    for(int i=0;i<n-1;i++)
    {   int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    ll sum1=0,sum2=0;
    vis[x]=true;
    dfs(x);
    sum1=sizes[y];
    memset(sizes,0,sizeof(sizes));
    memset(vis,0,sizeof(vis));
    vis[y]=true;
    dfs(y);
    sum2=sizes[x];
    //cout<<sum1<<" "<<sum2<<endl;
    ll ans=n*(n-1);
    ans=ans-sum1*sum2;
    cout<<ans<<endl;
    return 0;
}