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 ) ; } }