【codeforces-482div2-C】Kuro and Walking Route(DFS)

题目链接：http://codeforces.com/contest/979/problem/C

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 $\mathrm{nn, xx and yy (1\le n\le 3\cdot 1051\le n\le 3\cdot 105, 1\le x,y\le n1\le x,y\le n, x\ne yx\ne y)\; -\; the\; number\; of\; towns,\; index\; of\; the\; town\; Flowrisa\; and\; index\; of\; the\; town\; Beetopia,\; respectively.}$

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

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

input

3 1 3
1 2
2 3

output

5

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

题意

无向图，n个点，n-1条边，每两个点都可以到达，但是从依次经过u,v两点的道路不能走，问有多少个x->y可以到达

 思路

ans = 总路线条数 - u到v的路线数。u到v路线数 = u端的点数*v端的点数。判断点数用dfs。或者用SPFA记录u到v的所有点，再分别dfs u 和 v

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 3e5+4;
bool vis[N];
LL n, ans1 = 0, ans2 = 0, u, v, pre;
vector<int>V[N];
void dfs1(LL s, LL x)
{
    vis[s] = 1;
    if(s == v)
    {
        pre = x;
        return;
    }
    ans1++;
    for(LL i = 0; i < V[s].size(); i++)
    {
        LL k = V[s][i];
        if(vis[k]) continue;
        dfs1(k, s);
    }
}
void dfs2(int s)
{
    vis[s] = 1;
    if(s == u || s == v)
        return;
    ans2++;
    for(LL i = 0; i < V[s].size(); i++)
    {
        LL k = V[s][i];
        if(vis[k]) continue;
        dfs2(k);
    }
}
int main()
{
    LL a, b;
    scanf("%lld%lld%lld", &n, &u, &v);
    for(LL i = 1; i < n; i++)
    {
        scanf("%lld%lld", &a, &b);
        V[a].push_back(b);
        V[b].push_back(a);
    }
    dfs1(u, u);
    memset(vis, 0, sizeof vis);
    dfs2(pre);
    printf("%lld
", n*(n-1)-(ans1-ans2)*(n-ans1));
    return 0;
}