【POJ】1741 Tree 点分治

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传送门：【POJ】1741 Tree

题目分析：点分治第一题！

每次选择树的重心为根将树分治，每次分治加上所有深度和小于等于K的对数（O(N)可求），然后减去重心到儿子的边重叠的对数（在同一棵子树中的对数），再求出子树的根。

不断重复上述操作直到最后，即可求出答案。

论文题。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;typedef long long LL ;#define travel( 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 )const int MAXN = 10005 ;const int MAXE = 20005 ;struct Edge {int v , c ;Edge* next ;} E[MAXE] , *H[MAXN] , *edge ;int dep[MAXN] ;int siz[MAXN] ;int num[MAXN] ;int vis[MAXN] ;int tot_size ;int S[MAXN] ;int n , K ;int root ;int top ;int res ;void clear () {edge = E ;clr ( H , 0 ) ;clr ( vis , 0 ) ;num[0] = tot_size = n ;root = 0 ;}void addedge ( int u , int v , int c ) {edge -> v = v ;edge -> c = c ;edge -> next = H[u] ;H[u] = edge ++ ;}void get_root ( int u , int fa = 0 ) {//求树的重心siz[u] = 1 ;num[u] = 0 ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] && v != fa ) {get_root ( v , u ) ;siz[u] += siz[v] ;num[u] = max ( num[u] , siz[v] ) ;}}num[u] = max ( num[u] , tot_size - siz[u] ) ;if ( num[u] < num[root] ) root = u ;}void get_dep ( int u , int fa = 0 ) {//求到根的长度if ( dep[u] <= K ) S[top ++] = dep[u] ;siz[u] = 1 ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] && v != fa ) {dep[v] = dep[u] + e -> c ;get_dep ( v , u ) ;siz[u] += siz[v] ;}}}int get_num ( int u , int len ) {//得到对数top = 0 ;dep[u] = len ;get_dep ( u ) ;sort ( S , S + top ) ;int l = 0 , r = top - 1 , ans = 0 ;while ( l < r ) {if ( S[l] + S[r] <= K ) {ans += r - l ;++ l ;} else -- r ;}return ans ;}void dfs ( int u ) {//分治vis[u] = 1 ;res += get_num ( u , 0 ) ;travel ( e , H , u ) {int v = e -> v ;if ( !vis[v] ) {res -= get_num ( v , e -> c ) ;root = 0 ;tot_size = siz[v] ;get_root ( v ) ;dfs ( root ) ;}}}void solve () {int u , v , c ;clear () ;rep ( i , 1 , n ) {scanf ( "%d%d%d" , &u , &v , &c ) ;addedge ( u , v , c ) ;addedge ( v , u , c ) ;}res = 0 ;get_root ( 1 ) ;dfs ( root ) ;printf ( "%d\n" , res ) ;}int main () {while ( ~scanf ( "%d%d" , &n , &K ) && ( n || K ) ) solve () ;return 0 ;}

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