Poj2342 -- hdu1520 -- Anniversary party (tree DP Exercise 1), tree dp

Poj2342 -- hdu1520 -- Anniversary party (tree DP Exercise 1), tree dp

Anniversary party Time Limit:1000 MS Memory Limit:32768KB 64bit IO Format:% I64d & % I64uSubmit Status

Description

There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. the University has a hierarchical structure of employees. it means that the supervisor relation forms a tree rooted at the rector V. e. tretyakov. in order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. the personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. your task is to make a list of guests with the maximal possible sum of guests 'conviviality ratings.
 

Input

Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. each of the subsequent N lines contains the conviviality rating of the corresponding employee. conviviality rating is an integer number in a range from-128 to 127. after that go T lines that describe a supervisor relation tree. each line of the tree specification has the form:
L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line
0 0

Output

Output shoshould contain the maximal sum of guests ratings.
 

Sample Input

 711111111 32 36 47 44 53 50 0 

 

Sample Output

 5 

 

If a company wants to attend a banquet, the person who wants to attend the banquet must not have a direct leader relationship and give each person a happy value. l k Represents K as a direct leader of L, ask the maximum value of joy.

Build the company relationship into a tree from the largest boss to the dfs

Dp [I] [0] represents the maximum joy of a subtree (excluding I) rooted by the employee numbered I.

Dp [I] [1] represents the maximum joy of a subtree (including I) rooted by the employee numbered I.

The state transition equation is assumed that j is a subordinate of I.

Dp [I] [0] = max (dp [j] [0], dp [j] [1]);

Dp [I] [1] = max (dp [j] [0]) + c [I];

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;struct node{    int v ;    int next ;}edge[10000];int head[10000] , cnt ;int c[10000] ;int dp[10000][2] ;void add(int l,int k){    edge[cnt].v = l ;    edge[cnt].next = head[k] ;    head[k] = cnt++ ;}void dfs(int u){    if( head[u] == -1 )    {        dp[u][0] = 0 ;        dp[u][1] = c[u] ;        return ;    }    int i , v ;    for(i = head[u] , dp[u][1] = c[u] ; i != -1 ; i = edge[i].next)    {        v = edge[i].v ;        dfs(v) ;        dp[u][0] += max(dp[v][0],dp[v][1]) ;        dp[u][1] += dp[v][0] ;    }    return ;}int main(){    int n , i , j , l , k , num ;    while( scanf("%d", &n) != EOF )    {        memset(dp,0,sizeof(dp)) ;        memset(head,-1,sizeof(head)) ;        cnt = num = 0 ;        for(i = 1 ; i <= n ; i++)        {            scanf("%d", &c[i]) ;            num += i ;        }        while( scanf("%d %d", &l, &k) && (l+k != 0) )        {            num -= l ;            add(l,k) ;        }        dfs(num) ;        printf("%d\n", max( dp[num][0],dp[num][1] ) );    }    return 0 ;}