【opencup】XVI Open Cup named after E.V. Pankratiev. GP of SPB K.Captain Tarjan【树形dp】

题目大意：给一棵N个节点的树，你可以选择任意一个点作为根然后树链剖分（树链剖分的定义：如果这个点不是叶子，则其中恰好一条边为重边，其余均为轻边），剖分的方式由你定。然后接下来有M对(u,v)，定义(u,v)为树上节点u到节点v的最短路径。现在你要确定一个剖分方案，使得所有对(u,v)经过的轻边个数之和最小，经过多次算多次。输出个数之和。N,M∈[1,1e5]

sb题，100分钟后才过居然还在opencup的vp上拿了fb。

PS：经过多次算一次已经最大个数之和均是一种做法。

#include <bits/stdc++.h>
using namespace std ;

typedef long long LL ;
typedef pair < int , int > pii ;

#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 100005 ;

vector < pii > G[MAXN] ;
LL dp[MAXN] , dp2[MAXN] , maxc[MAXN] ;
LL Ldp[MAXN] , Rdp[MAXN] , Lmaxc[MAXN] , Rmaxc[MAXN] ;
int c[MAXN] ;
int siz[MAXN] ;
int son[MAXN] ;
int pre[MAXN] ;
int top[MAXN] ;
int dep[MAXN] ;
int n , m ;
LL ans ;

void predfs ( int u , int f ) {
siz[u] = 1 ;
son[u] = 0 ;
for ( int i = 0 ; i < G[u].size () ; ++ i ) {
int v = G[u][i].first ;
if ( v == f ) continue ;
pre[v] = u ;
dep[v] = dep[u] + 1 ;
predfs ( v , u ) ;
siz[u] += siz[v] ;
if ( siz[v] > siz[son[u]] ) son[u] = v ;
}
}

void rebuild ( int u , int top_element ) {
top[u] = top_element ;
if ( son[u] ) rebuild ( son[u] , top_element ) ;
for ( int i = 0 ; i < G[u].size () ; ++ i ) {
int v = G[u][i].first ;
if ( v != pre[u] && v != son[u] ) rebuild ( v , v ) ;
}
}

int get_lca ( int x , int y ) {
while ( top[x] != top[y] ) {
if ( dep[top[x]] > dep[top[y]] ) x = pre[top[x]] ;
else y = pre[top[y]] ;
}
return dep[x] < dep[y] ? x : y ;
}

void dfs1 ( int u ) {
for ( int i = 0 ; i < G[u].size () ; ++ i ) {
int v = G[u][i].first ;
if ( v == pre[u] ) continue ;
dfs1 ( v ) ;
G[u][i].second = c[v] ;
c[u] += c[v] ;
}
}

void dfs2 ( int u ) {
dp[u] = maxc[u] = 0 ;
for ( int i = 0 ; i < G[u].size () ; ++ i ) {
int v = G[u][i].first ;
if ( v == pre[u] ) continue ;
dfs2 ( v ) ;
dp[u] += dp[v] + G[u][i].second ;
maxc[u] = max ( maxc[u] , G[u][i].second + 0LL ) ;
}
dp[u] -= maxc[u] ;
}

void dfs3 ( int u , int cost ) {
int n = G[u].size () ;
Ldp[0] = 0 ;
Rdp[n - 1] = 0 ;
Lmaxc[0] = 0 ;
Rmaxc[n - 1] = 0 ;
for ( int i = 1 ; i < n ; ++ i ) {
int v = G[u][i - 1].first ;
if ( v == pre[u] ) {
Ldp[i] = Ldp[i - 1] ;
Lmaxc[i] = Lmaxc[i - 1] ;
} else {
Ldp[i] = Ldp[i - 1] + dp[v] + G[u][i - 1].second ;
Lmaxc[i] = max ( Lmaxc[i - 1] , G[u][i - 1].second + 0LL ) ;
}
}
for ( int i = n - 2 ; i >= 0 ; -- i ) {
int v = G[u][i + 1].first ;
if ( v == pre[u] ) {
Rdp[i] = Rdp[i + 1] ;
Rmaxc[i] = Rmaxc[i + 1] ;
} else {
Rdp[i] = Rdp[i + 1] + dp[v] + G[u][i + 1].second ;
Rmaxc[i] = max ( Rmaxc[i + 1] , G[u][i + 1].second + 0LL ) ;
}
}
for ( int i = 0 ; i < n ; ++ i ) {
int v = G[u][i].first ;
if ( v == pre[u] ) continue ;
dp2[v] = dp2[u] + Ldp[i] + Rdp[i] + cost - max ( max ( Lmaxc[i] , Rmaxc[i] ) , cost + 0LL ) ;
ans = min ( ans , dp2[v] + dp[v] + maxc[v] + G[u][i].second - max ( maxc[v] , 0LL + G[u][i].second ) ) ;
}
for ( int i = 0 ; i < n ; ++ i ) {
int v = G[u][i].first ;
if ( v == pre[u] ) continue ;
dfs3 ( v , G[u][i].second ) ;
}
}

void solve () {
for ( int i = 1 ; i <= n ; ++ i ) {
G[i].clear () ;
c[i] = 0 ;
}
for ( int i = 1 ; i < n ; ++ i ) {
int u , v ;
scanf ( "%d%d" , &u , &v ) ;
G[u].push_back ( pii ( v , 0 ) ) ;
G[v].push_back ( pii ( u , 0 ) ) ;
}
predfs ( 1 , 1 ) ;
rebuild ( 1 , 1 ) ;
for ( int i = 0 ; i < m ; ++ i ) {
int u , v ;
scanf ( "%d%d" , &u , &v ) ;
c[u] ++ ;
c[v] ++ ;
int lca = get_lca ( u , v ) ;
c[lca] -= 2 ;
}
dp2[1] = 0 ;
dfs1 ( 1 ) ;
dfs2 ( 1 ) ;
ans = dp[1] ;
dfs3 ( 1 , 0 ) ;
printf ( "%lld\n" , ans ) ;
}

int main () {
freopen ( "treepaths.in" , "r" , stdin ) ;
freopen ( "treepaths.out" , "w" , stdout ) ;
while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;
return 0 ;
}