HDU1520 Anniversary party 树形DP

               

Problem 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 should contain the maximal sum of guests' ratings.

 

Sample Input

711111111 32 36 47 44 53 50 0

 

Sample Output

5

 

 

这道题的题意与POJ2342是一样的，具体在这http://blog.youkuaiyun.com/libin56842/article/details/9814455

但是，将POJ的代码贴上去是WA，加上一个n为0退出的条件是TLE

然后我还发现，如果调用结构体内的函数用时109ms

如果用外在函数，同样超时

看来调用结构体内的函数比较节省时间

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{    int child,father,brother,present,not_present;    int max()//结构体内的函数调用时耗时少    {        return present>not_present?present:not_present;    }    void init()    {        child=father=brother=not_present=0;    }}tree[6005];void dfs(int root){    int son = tree[root].child;    while(son)    {        dfs(son);        tree[root].present+=tree[son].not_present;        tree[root].not_present+=tree[son].max();        son = tree[son].brother;    }}int main(){    int n,i,j,k,l;    while(~scanf("%d",&n)&&n)    {        for(i = 1;i<=n;i++)        {            scanf("%d",&tree[i].present);            tree[i].init();        }        while(~scanf("%d%d",&l,&k),l+k)        {            tree[l].father = k;            tree[l].brother = tree[k].child;            tree[k].child = l;        }        for(i = 1;i<=n;i++)        {            if(!tree[i].father)            {                dfs(i);                printf("%d\n",tree[i].max());                break;            }        }    }    return 0;}

 

           

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