学步园

传送门：【SPOJ】913 Query on a tree II

题目分析：LCA倍增算法的应用，求两点间距离是没问题的，重要的是求路径上第k个点，首先LCA一次，看这个点是在x到祖先的路径上还是y到祖先的位置上，然后继续用类似二分的思想求这个点即可。

代码如下：

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

typedef long long LL ;

#define logn( i , a , b ) for ( int i = ( a ) ; ( 1 << i ) <= ( b ) ; ++ i )
#define repp( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 2 , n
#define mid ( ( l + r ) >> 1 )

const int LOG = 20 ;
const int MAXN = 10005 ;
const int MAXE = 20005 ;

struct Edge {
	int v , c ;
	Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;

int anc[MAXN][LOG] ;
int dist[MAXN] ;
int pre[MAXN] ;
int dep[MAXN] ;
int n ;

void clear () {
	edge = E ;
	clr ( H , 0 ) ;
	clr ( dep , 0 ) ;
	clr ( dist , 0 ) ;
}

void addedge ( int u , int v , int c ) {
	edge -> v = v ;
	edge -> c = c ;
	edge -> next = H[u] ;
	H[u] = edge ++ ;
	edge -> v = u ;
	edge -> c = c ;
	edge -> next = H[v] ;
	H[v] = edge ++ ;
}

void dfs ( int u , int fa = -1 ) {
	pre[u] = fa ;
	repp ( e , H , u ) {
		int v = e -> v ;
		if ( v == fa ) continue ;
		dist[v] = dist[u] + e -> c ;
		dep[v] = dep[u] + 1 ;
		dfs ( v , u ) ;
	}
}

void preprocess () {
	FOR ( i , 1 , n ) anc[i][0] = pre[i] ;
	FOR ( i , 1 , n ) logn ( j , 1 , n ) anc[i][j] = -1 ;
	logn ( j , 1 , n ) FOR ( i , 1 , n ) if ( ~anc[i][j - 1] ) anc[i][j] = anc[anc[i][j - 1]][j - 1] ;
}

int lca ( int x , int y ) {
	if ( dep[x] < dep[y] ) swap ( x , y ) ;
	int log = 0 ;
	logn ( i , 1 , dep[x] ) ++ log ;
	rev ( i , log , 0 ) if ( dep[x] - ( 1 << i ) >= dep[y] ) x = anc[x][i] ;
	if ( x == y ) return x ;
	rev ( i , log , 0 ) if ( ~anc[x][i] && anc[x][i] != anc[y][i] ) {
		x = anc[x][i] ;
		y = anc[y][i] ;
	}
	return pre[x] ;
}

int kth ( int x , int y , int k ) {
	int cp = lca ( x , y ) ;
	if ( dep[x] - dep[cp] + 1 >= k ) {
		int deep = dep[x] - k + 1 , log = 0 ;
		logn ( i , 1 , dep[x] ) ++ log ;
		rev ( i , log , 0 ) if ( dep[x] - ( 1 << i ) >= deep ) x = anc[x][i] ;
		return x ;
	} else {
		int deep = k - ( dep[x] - dep[cp] ) + dep[cp] - 1 , log = 0 ;
		logn ( i , 1 , dep[y] ) ++ log ;
		rev ( i , log , 0 ) if ( dep[y] - ( 1 << i ) >= deep ) y = anc[y][i] ;
		return y ;
	}
}

void solve () {
	char s[10] ;
	int x , y , c ;
	clear () ;
	scanf ( "%d" , &n ) ;
	rep ( i , 1 , n ) {
		scanf ( "%d%d%d" , &x , &y , &c ) ;
		addedge ( x , y , c ) ;
	}
	dfs ( 1 ) ;
	preprocess () ;
	while ( scanf ( "%s" , s ) && s[1] != 'O' ) {
		if ( s[0] == 'D' ) {
			scanf ( "%d%d" , &x , &y ) ;
			int cp = lca ( x , y ) ;
			printf ( "%d\n" , dist[x] + dist[y] - 2 * dist[cp] ) ;
		} else {
			scanf ( "%d%d%d" , &x , &y , &c ) ;
			printf ( "%d\n" , kth ( x , y , c ) ) ;
		}
	}
}

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) solve () ;
	return 0 ;
}