BZOJ 2152: 聪聪可可【树形dp

……煞笔树形dp看错题调了一年怎么办QAQ

…直接上代码了没什么好说的

码风越来越丑我也很绝望啊【

#include
#define MAXN 20005
using namespace std ;	int n ;
inline int read() {
	register int ch = getchar() ;
	while (!isdigit(ch))	ch = getchar() ;
	register int rtn = 0 ;
	while (isdigit(ch))	rtn = rtn*10 + ch - '0' , ch = getchar() ;
	return rtn ;
}

int f[MAXN][3] ;
int T[3] , T1[3] ;
int ans[3] ;
// f[] can be linked up

struct t1 {
	int to, nxt, lth ;
	t1(){}
	t1(int to, int nxt, int lth) : to(to), nxt(nxt), lth(lth) {}
}edge[MAXN<<1] ;	int cnt_edge = 0 ;
int fst[MAXN] ;
inline void addedge(int x, int y, int lth) {
	edge[++cnt_edge] = t1(y, fst[x], lth) ;
	fst[x] = cnt_edge ;
	edge[++cnt_edge] = t1(x, fst[y], lth) ;
	fst[y] = cnt_edge ;
}

void dfs( int now , int fth ) {
	for ( int tmp = fst[now] ; tmp ; tmp = edge[tmp].nxt ) {
		int aim = edge[tmp].to ;

		if (aim == fth)	continue ;
		dfs(aim,now) ;

		for (int i=0; i^3; ++i)	T[ (i+edge[tmp].lth) %3 ] = f[aim][i] ;
		for (int i=0; i^3; ++i)	f[aim][i] = T[i] ;
		
		T1[0] = T1[1] = T1[2] = 0 ;
		
		for (int i=0; i^3; ++i) {
			for (int j=0; j^3; ++j) {
				ans[i] += f[aim][j] * f[now][ (i+3-j)%3 ] ; 
			}
			T1[i] += f[aim][i] ;
			ans[i] += f[aim][i] ;
		}
		
		f[now][0] += T1[0] , f[now][1] += T1[1] , f[now][2] += T1[2] ;

//		printf("aim = %d\n",aim) ;
//		printf("now = %d\n",now) ;
//		printf("%d %d %d\n",f[now][0] , f[now][1] , f[now][2] ) ;
//		printf("ans = %d %d %d\n\n",ans[0] , ans[1] , ans[2] ) ;
	}
	++f[now][0] ;
}

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

int main() {
//	freopen("1.in","r",stdin) ;

	n = read() ;
	for (int i=1; i^n; ++i) {
		int x = read() , y = read() , lth = read() % 3 ;
		addedge(x,y,lth) ;
	}

	dfs(1,0) ;

	int up = ans[0] , down = ans[0] + ans[1] + ans[2] ;
	up = ( up << 1 ) + n , down = ( down << 1 ) + n ;
	int dd = gcd(up, down) ;
	up /= dd , down /= dd ;
	
	printf("%d/%d\n",up,down) ;

	return 0 ;
}