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

Posted 2020-11-07 灬从此以后灬

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

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 $(u,v) he\; can\; take\; as\; his\; route.\; Since\; he’s\; not\; really\; bright,\; he\; asked\; you\; to\; help\; him\; with\; this\; problem.$

Input

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

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

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:

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

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:

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

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+50 ,mod = 998244353,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]=1;
    siz[k]=1;
    fa[k]=f;
    for(int i=0;i<g[k].size();i++)
    {
        int u=g[k][i];
        if(vis[u]==0&&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]==0)
        return 1;
    else
        return dfs2(fa[k1],k2);
}
int main()
{
    cin>>n>>x>>y;
    for(int i=0;i<n-1;i++)
    {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    memset(siz,0,sizeof(siz));
    memset(vis,0,sizeof(vis));
    dfs1(1,0);
    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==1)
        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-1)-ans<<endl;
}