poj2486 Apple Tree （树形dp）

题意：有一颗苹果树，树上的u节点上有num[u]个苹果，树根为1号节点，囧king从根开始走，没走到一个节点就把接点上的苹果吃光，问囧king在不超过k步的情况下最多吃多少个苹果。

解题思路：处理出两个dp数组，f1[u][i]表示在不超过i步的情况下，从u节点开始，往下吃，吃完后回到u节点，最多能吃多少苹果。f2[u][i]表示在不超过i步的情况下，从u节点开始往下吃，最多能吃多少苹果。

#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 ) ;
	}
}