題意:有一顆蘋果樹,樹上的u節點上有num[u]個蘋果,樹根為1號節點,囧king從根開始走,沒走到一個節點就把接點上的蘋果吃光,問囧king在不超過k步的情況下最多吃多少個蘋果。
題意:有一顆蘋果樹,樹上的u節點上有num[u]個蘋果,樹根為1號節點,囧king從根開始走,沒走到一個節點就把接點上的蘋果吃光,問囧king在不超過k步的情況下最多吃多少個蘋果。解題思路:處理出兩個dp數組,f1[u][i]表示在不超過i步的情況下,從u節點開始,往下吃,吃完後回到u節點,最多能吃多少蘋果。f2[u][i]表示在不超過i步的情況下,從u節點開始往下吃,最多能吃多少蘋果。
解題思路:處理出兩個dp數組,f1[u][i]表示在不超過i步的情況下,從u節點開始,往下吃,吃完後回到u節點,最多能吃多少蘋果。f2[u][i]表示在不超過i步的情況下,從u節點開始往下吃,最多能吃多少蘋果。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std ;
const int maxn = 111111 ;
int max ( int a , int b ) { return a > b ? a : b ; }
int min ( int a , int b ) { return a < b ? a : b ; }
struct Edge
{
int to , next ;
} edge[maxn<<1];
int head[maxn] , tot , n , m ;
int f1[111][2222] , f2[111][222] , num[maxn] , ans , dis[1111] ;
void new_edge ( int a , int b )
{
edge[tot].to = b ;
edge[tot].next = head[a] ;
head[a] = tot ++ ;
edge[tot].to = a ;
edge[tot].next = head[b] ;
head[b] = tot ++ ;
}
int c[111][222] , d[maxn] , l ;
void dfs ( int u , int fa , int *f )
{
int i , j , k ;
if ( u != 1 )
{
for ( i = dis[u] ; i <= m ; i ++ )
ans = max ( ans , f2[u][m-i] + f[i] ) ;
}
int fuck[222] ;
for ( i = head[u] ; i != -1 ; i = edge[i].next )
{
int v = edge[i].to ;
if ( v == fa ) continue ;
for ( j = 0 ; j <= m ; j ++ ) d[j] = 0 ;
int t = 0 ;
for ( j = head[u] ; j != -1 ; j = edge[j].next )
{
if ( edge[j].to == v || edge[j].to == fa ) continue;
t ++ ;
for ( k = 0 ; k <= m ; k ++ )
c[t][k] = f1[edge[j].to][k] ;
}
for ( j = 1 ; j <= t ; j ++ )
for ( k = m ; k >= 0 ; k -- )
for ( l = 0 ; l + 2 <= k ; l ++ )
d[k] = max ( d[k] , d[k-l-2] + c[j][l] ) ;
for ( j = 0 ; j <= m ; j ++ ) fuck[j] = 0 ;
for ( j = dis[u] ; j < m ; j ++ ) fuck[j+1] = f[j] ;
for ( j = dis[u] ; j <= m ; j ++ )
for ( k = 0 ; k <= m ; k ++ )
if ( j + k + 1 <= m )
fuck[j+k+1] = max ( fuck[j+k+1] , f[j] + d[k] + num[u] ) ;
dfs ( v , u , fuck ) ;
}
}
void cal ( int u , int fa )
{
int i , j , k ;
for ( i = head[u] ; i != -1 ; i = edge[i].next )
{
int v = edge[i].to ;
if ( v == fa ) continue ;
dis[v] = dis[u] + 1 ;
cal ( v , u ) ;
for ( k = m ; k >= 0 ; k -- )
for ( j = 0 ; j + 2 <= k ; j ++ )
f1[u][k] = max ( f1[u][k] , f1[u][k-j-2] + f1[v][j] ) ;
for ( k = 1 ; k <= m ; k ++ ) f2[u][k] = max ( f2[u][k] , f2[v][k-1] ) ;
for ( j = 0 ; j <= m ; j ++ ) d[j] = 0 ;
int t = 0 ;
for ( j = head[u] ; j != -1 ; j = edge[j].next )
{
if ( edge[j].to == v || edge[j].to == fa ) continue ;
t ++ ;
for ( k = 0 ; k <= m ; k ++ )
c[t][k] = f1[edge[j].to][k] ;
}
for ( j = 1 ; j <= t ; j ++ )
for ( k = m ; k >= 0 ; k -- )
for ( l = 0 ; l + 2 <= k ; l ++ )
d[k] = max ( d[k] , d[k-l-2] + c[j][l] ) ;
for ( j = 0 ; j <= m ; j ++ )
for ( k = 0 ; k + 1 <= j ; k ++ )
f2[u][j] = max ( f2[u][j] , f2[v][k] + d[j-k-1] ) ;
}
for ( i = 0 ; i <= m ; i ++ ) f1[u][i] += num[u] , f2[u][i] += num[u] ;
for ( i = 1 ; i <= m ; i ++ ) f1[u][i] = max ( f1[u][i] , f1[u][i-1] ) , f2[u][i] = max ( f2[u][i] , f2[u][i-1] ) ;
}
void init ()
{
int i ;
memset ( head , -1 , sizeof ( head ) ) ;
memset ( f1 , 0 , sizeof ( f1 ) ) ;
memset ( f2 , 0 , sizeof ( f2 ) ) ;
memset ( dis , 0 , sizeof ( dis ) ) ;
tot = ans = 0 ;
}
int f[222] ;
int main ()
{
int i , j , k , a , b ;
while ( scanf ( "%d%d" , &n , &m ) != EOF )
{
init () ;
for ( i = 1 ; i <= n ; i ++ ) scanf ( "%d" , &num[i] ) ;
for ( i = 1 ; i < n ; i ++ )
{
scanf ( "%d%d" , &a , &b ) ;
new_edge ( a , b ) ;
}
cal ( 1 , 0 ) ;
ans = f2[1][m] ;
for ( i = 0 ; i <= m ; i ++ ) f[i] = 0 ;
dfs ( 1 , 0 , f ) ;
printf ( "%d\n" , ans ) ;
}
}