这道题用 LCA 就可以水过去 , 但是我太弱了 QAQ 倍增写LCA总是写残...于是就写了树链剖分...
其实也不难写 , 线段树也不用用到 , 自己YY一下然后搞一搞就过了...速度还挺快的好像= = #9
----------------------------------------------------------------------------------
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define rep( i , n ) for( int i = 0 ; i < n ; i++ )
#define clr( x , c ) memset( x , c , sizeof( x ) )
#define REP( x ) for( edge* e = head[ x ] ; e ; e = e -> next )
using namespace std;
const int maxn = 500000 + 5;
struct edge {
int to;
edge* next;
};
edge* pt , EDGE[ maxn << 1 ];
edge* head[ maxn ];
void edge_init() {
pt = EDGE;
clr( head , 0 );
}
void add( int u , int v ) {
pt -> to = v;
pt -> next = head[ u ];
head[ u ] = pt++;
}
#define add_edge( u , v ) add( u , v ) , add( v , u )
int fa[ maxn ] , top[ maxn ] , size[ maxn ] , son[ maxn ] , dep[ maxn ];
int id[ maxn ] , _id[ maxn ] , id_cnt = 0 , TOP;
void dfs( int x ) {
size[ x ] = 1;
son[ x ] = -1;
REP( x ) {
int to = e -> to;
if( to == fa[ x ] ) continue;
dep[ to ] = dep[ x ] + 1;
fa[ to ] = x;
dfs( to );
size[ x ] += size[ to ];
if( son[ x ] == - 1 || size[ to ] > size[ son[ x ] ] )
son[ x ] = to;
}
}
void DFS( int x ) {
top[ x ] = TOP;
id[ x ] = ++id_cnt;
_id[ id_cnt ] = x;
if( son[ x ] != -1 )
DFS( son[ x ] );
REP( x ) if( id[ e -> to ] < 0 )
DFS( TOP = e -> to );
}
void DFS_init() {
clr( id , -1 );
dfs( dep[ 0 ] = 0 );
DFS( TOP = 0 );
}
int Q_LCA( int x , int y ) {
while( top[ x ] != top[ y ] ) {
if( dep[ top[ x ] ] < dep[ top[ y ] ] )
swap( x , y );
x = fa[ top[ x ] ];
}
return dep[ x ] < dep[ y ] ? x : y;
}
int Q_dist( int x , int y ) {
int res = 0;
while( top[ x ] != top[ y ] ) {
if( dep[ top[ x ] ] < dep[ top[ y ] ] )
swap( x , y );
res += id[ x ] - id[ top[ x ] ];
x = fa[ top[ x ] ];
res++;
}
if( dep[ x ] < dep[ y ] )
swap( x , y );
return res + id[ x ] - id[ y ];
}
int ans[ 2 ];
void work( int x , int y , int z ) {
int LCA[ 3 ] = { Q_LCA( x , y ) , Q_LCA( x , z ) , Q_LCA( y , z ) };
if( LCA[ 0 ] == LCA[ 1 ] )
ans[ 0 ] = LCA[ 2 ];
else if( LCA[ 0 ] == LCA[ 2 ] )
ans[ 0 ] = LCA[ 1 ];
else
ans[ 0 ] = LCA[ 0 ];
ans[ 1 ] = Q_dist( x , ans[ 0 ] ) + Q_dist( y , ans[ 0 ] ) + Q_dist( z , ans[ 0 ] );
}
inline int read() {
int res = 0;
char c = getchar();
while( ! isdigit( c ) ) c = getchar();
while( isdigit( c ) ) {
res = res * 10 + c - ‘0‘;
c = getchar();
}
return res;
}
int main() {
freopen( "test.in" , "r" , stdin );
int n = read() , m = read();
edge_init();
rep( i , n - 1 ) {
int u = read() , v = read();
u-- , v--;
add_edge( u , v );
}
DFS_init();
while( m-- ) {
int x = read() , y = read() , z = read();
x-- , y-- , z--;
work( x , y , z );
printf( "%d %d\n" , ans[ 0 ] + 1 , ans[ 1 ] );
}
return 0;
}
----------------------------------------------------------------------------------
1787: [Ahoi2008]Meet 紧急集合
Time Limit: 20 Sec Memory Limit: 162 MB
Submit: 1542 Solved: 683
[Submit][Status][Discuss]
Description
Input
Output
Sample Input
6 4
1 2
2 3
2 4
4 5
5 6
4 5 6
6 3 1
2 4 4
6 6 6
Sample Output
5 2
2 5
4 1
6 0
HINT
Source
BZOJ 1787: [Ahoi2008]Meet 紧急集合( 树链剖分 )
原文：http://www.cnblogs.com/JSZX11556/p/4574766.html
