Wannafly挑战赛1 A Treepath LCA

https://www.nowcoder.com/acm/contest/15/A

观察到路径x->y的长度的奇偶性只与x到根的长度的奇偶性以及y到根的长度的奇偶性有关（令dist[x]表示x到根的距离，x到y的距离为dist[x]+dist[y]-2dist[lca]，最后一项对奇偶性没影响，所以真正有用的信息包含在dist[x]和dist[y]里，不难想到当这两个数奇偶性相同时路径长度为偶数）

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.
///             __.'              ~.   .~              `.__
///           .'//                  \./                  \\`.
///        .'//                     |                     \\`.
///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.
///     .'//.-"                 `-.  |  .-'                 "-.\\`.
///   .'//______.============-..   \ | /   ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//void readString(string &s)
//{
//	static char str[maxn];
//	scanf("%s", str);
//	s = str;
//}

int n;
int num[maxn];
vector<int>G[maxn];
bool vis[maxn];
void dfs(int u,int cnt)
{
	num[cnt]++;
	vis[u]=1;
	for(int i:G[u])
		if(!vis[i])
			dfs(i,cnt^1);
}
	
void solve()
{
	s_1(n);
	FOR(1,n-1,i)
	{
		int u,v;
		s_2(u,v);
		G[u].pb(v);
		G[v].pb(u);
	}
	dfs(1,0);
	print((LL)num[0]*(num[0]-1ll)/2+(LL)num[1]*(num[1]-1ll)/2);
}

int main()
{
    //freopen( "1.in" , "r" , stdin );
    //freopen( "1.out" , "w" , stdout );
    int t=1;
    //init();
    //s_1(t);
    for(int cas=1;cas<=t;cas++)
    {
        //printf("Case #%d: ",cas);
        solve();
    }
}