【ZJU】3863 Paths on the Tree 【浙大2015年4月校赛D题】 树分治

传送门：【ZJU】3863 Paths on the Tree

题意：给一棵树，问树上有多少个路径对有不超过K个公共节点的，路径a->b和b->a等价，路径对（A，B）和（B，A）只有当A和B是同一条路径时相同。

分析：反过来考虑，考虑有超过K+1个公共节点的路径对数。我们考虑重叠的路径部分，这个可以用树分治来搞，然后路径对的两端延伸出去的部分不重叠，我们要预处理出这个部分。最后就是当前枚举的子树和之前子树乘一乘。

trick：路径对总数是会爆long long的，要用unsigned long long。

后记：太懒了。。写不动题解啊，粗略的写了下思路，细节可以自己想想，还不理解可以QQ找我。。

代码如下：

#include <stdio.h>#include <string.h>#include <string>#include <math.h>#include <map>#include <algorithm>using namespace std ;typedef long long LL ;typedef unsigned long long ULL ;#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 = 400005 ;struct Edge {int v , c , n ;Edge () {}Edge ( int v , int n ) : v ( v ) , n ( n ) {}} ;struct Node {int d ;ULL x ;Node () {}Node ( int d , ULL x ) : d ( d ) , x ( x ) {}} ;Edge E[MAXE] ;int H[MAXN] , cntE ;int vis[MAXN] , Time ;int siz[MAXN] ;int pre[MAXN] ;int dep[MAXN] ;int max_dis , tree_size ;int Q[MAXN] , head , tail ;Node S[MAXN] ;int top ;ULL c[MAXN] , cc[MAXN] ;int n , k ;ULL ans , tmpc0 ;void clear () {ans = 0 ;cntE = 0 ;++ Time ;clr ( H , -1 ) ;}void addedge ( int u , int v ) {E[cntE] = Edge ( v , H[u] ) ;H[u] = cntE ++ ;}int get_root ( int s ) {head = tail = 0 ;Q[tail ++] = s ;dep[s] = 0 ;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 ) continue ;pre[v] = u ;dep[v] = dep[u] + 1 ;Q[tail ++] = v ;}}max_dis = dep[Q[tail - 1]] ;int root = s , root_siz = MAXN , max_siz = tail ;tree_size = tail ;while ( head ) {int u = Q[-- head] ;int cnt = 0 ;siz[u] = 1 ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( v == pre[u] || vis[v] == Time ) continue ;siz[u] += siz[v] ;if ( siz[v] > cnt ) cnt = siz[v] ;}cnt = max ( cnt , max_siz - siz[u] ) ;if ( cnt < root_siz ) {root_siz = cnt ;root = u ;}}return root ;}void calc ( int s , int s_size , int root ) {head = tail = 0 ;Q[tail ++] = s ;dep[s] = 1 ;pre[s] = root ;top = 0 ;ULL c0 = tmpc0 - ( ULL ) 2 * s_size * ( n - s_size ) ;while ( head != tail ) {int u = Q[head ++] ;ULL x = 1 , y = 0 ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v , c = E[i].c ;if ( v == pre[u] ) continue ;x += E[i].c ;y += ( ULL ) E[i].c * E[i].c ;}//printf ( "%I64u %I64u %d %d\n" , x , y , dep[u] , u ) ;S[top ++] = Node ( dep[u] , x * x - y ) ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( v == pre[u] || vis[v] == Time ) continue ;pre[v] = u ;dep[v] = dep[u] + 1 ;Q[tail ++] = v ;}}rep ( i , 0 , top ) {int idx = max ( 0 , k - S[i].d - 1 ) ;if ( idx > max_dis ) continue ;if ( !idx ) ans += c0 * S[i].x ;ans += cc[idx] * S[i].x ;}rep ( i , 0 , top ) c[S[i].d] += S[i].x ;cc[max_dis] = c[max_dis] ;rev ( i , max_dis , 1 ) cc[i - 1] = cc[i] + c[i - 1] ;}void dfs ( int u ) {int root = get_root ( u ) ;if ( tree_size < k ) return ;//no satisfied pathvis[root] = Time ;memset ( c , 0 , sizeof ( c[0] ) * ( max_dis + 1 ) ) ;memset ( cc , 0 , sizeof ( cc[0] ) * ( max_dis + 1 ) ) ;//For ( i , 0 , max_dis ) c[i] = cc[i] = 0 ;ULL y = 0 ;for ( int i = H[root] ; ~i ; i = E[i].n ) y += ( ULL ) E[i].c * E[i].c ;tmpc0 = ( ULL ) n * n - y ;for ( int i = H[root] ; ~i ; i = E[i].n ) {int v = E[i].v ;if( vis[v] == Time ) continue ;calc ( v , E[i].c , root ) ;}for ( int i = H[root] ; ~i ; i = E[i].n ) if ( vis[E[i].v] != Time ) dfs ( E[i].v ) ;}void pre_dfs ( int u , int f ) {siz[u] = 1 ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( v == f ) continue ;pre_dfs ( v , u ) ;E[i].c = siz[v] ;siz[u] += siz[v] ;}for ( int i = H[u] ; ~i ; i = E[i].n ) if ( E[i].v == f ) {E[i].c = n - siz[u] ;break ;}}void solve () {int u , v ;clear () ;scanf ( "%d%d" , &n , &k ) ;++ k ;rep ( i , 1 , n ) {scanf ( "%d%d" , &u , &v ) ;addedge ( u , v ) ;addedge ( v , u ) ;}ULL x = ( ULL ) n * ( n + 1 ) / 2 ;pre_dfs ( 1 , 0 ) ;dfs ( 1 ) ;printf ( "%llu\n" , x * x - ans ) ;}int main () {int T ;Time = 0 ;clr ( vis , 0 ) ;scanf ( "%d" , &T ) ;For ( i , 1 , T ) solve () ;return 0 ;}