【BZOJ】2152: 聪聪可可 点分治

传送门：【BZOJ】2152: 聪聪可可

题目分析：记录权值和%3的路径的个数。。。然后去重。。没了。。

代码如下：

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;
const int MAXE = 200005 ;

struct Edge {
	int v , c , n ;
	Edge () {}
	Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}
} ;

Edge E[MAXE] ;
int H[MAXN] , cntE ;
int dis[MAXN] ;
int vis[MAXN] , Time ;
int pre[MAXN] ;
int siz[MAXN] ;
int dp[MAXN] ;
int d[3] ;
int Q[MAXN] , head , tail ;
int ans ;
int n ;

void clear () {
	cntE = 0 ;
	clr ( H , -1 ) ;
	Time ++ ;
	ans = 0 ;
}

void addedge ( int u , int v , int c ) {
	E[cntE] = Edge ( v , c , H[u] ) ;
	H[u] = cntE ++ ;
}

int gcd ( int a , int b ) {
	return b ? gcd ( b , a % b ) : a ;
}

int get_root ( int s ) {
	head = tail = 0 ;
	Q[tail ++] = s ;
	pre[s] = 0 ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v ;
			if ( v != pre[u] && vis[v] != Time ) {
				pre[v] = u ;
				Q[tail ++] = v ;
			}
		}
	}
	int tot_size = tail ;
	int root = s , max_size = tail ;
	while ( tail ) {
		int u = Q[-- tail] ;
		siz[u] = 1 ;
		int cnt = 0 ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v ;
			if ( v != pre[u] && vis[v] != Time ) {
				siz[u] += siz[v] ;
				cnt = max ( cnt , siz[v] ) ;
			}
		}
		cnt = max ( cnt , tot_size - siz[u] ) ;
		if ( cnt < max_size ) {
			max_size = cnt ;
			root = u ;
		}
	}
	return root ;
}

int get_ans ( int s , int init_dis ) {
	head = tail = 0 ;
	Q[tail ++] = s ;
	pre[s] = 0 ;
	dis[s] = init_dis % 3 ;
	d[0] = d[1] = d[2] = 0 ;
	while ( head != tail ) {
		int u = Q[head ++] ;
		d[dis[u]] ++ ;
		for ( int i = H[u] ; ~i ; i = E[i].n ) {
			int v = E[i].v ;
			if ( v != pre[u] && vis[v] != Time ) {
				pre[v] = u ;
				Q[tail ++] = v ;
				dis[v] = ( dis[u] + E[i].c ) % 3 ;
			}
		}
	}
	return d[1] * d[2] * 2 + d[0] * d[0] ;
}

void divide ( int u ) {
	int root = get_root ( u ) ;
	vis[root] = Time ;
	ans += get_ans ( root , 0 ) ;
	for ( int i = H[root] ; ~i ; i = E[i].n ) {
		int v = E[i].v ;
		if ( vis[v] != Time ) {
			ans -= get_ans ( v , E[i].c ) ;
			divide ( v ) ;
		}
	}
	vis[root] = 0 ;
}

void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void solve () {
	int u , v , c ;
	clear () ;
	rep ( i , 1 , n ) {
		scanf ( u ) ;
		scanf ( v ) ;
		scanf ( c ) ;
		addedge ( u , v , c % 3 ) ;
		addedge ( v , u , c % 3 ) ;
	}
	divide ( 1 ) ;
	//printf ( "%lld %lld\n" , ans , ( LL ) n * n ) ;
	int __gcd = gcd ( n * n , ans ) ;
	printf ( "%d/%d\n" , ans / __gcd , n * n / __gcd ) ;
}

int main () {
	clr ( vis , 0 ) ;
	Time = 0 ;
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}