HDU-1520-Anniversary party

                                     Anniversary party

                                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                Total Submission(s): 13361    Accepted Submission(s): 5314

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

这道题是入门题吧。

题意：上司和员工都有各自的欢乐值，然后上司与员工只能来一个，求最大的欢乐值。

就两种情况，当前人员来或者不来，dp[i][0]与dp[i][1]分别表示I点表示的人来或者不来，所以dp[i][0]=max(dp[son][0],dp[son][1]),即i点所代表的人不来的最大欢乐值取决于儿子节点的员工来或者不来，然后先找出根节点，从根节点开始搜索，层层更新，求出最终结果。

代码：

#include<stdio.h>#include<iostream>#include<math.h>#include<string.h>using namespace std;int n;int dp[6005][2],father[6005];void dfs(int p){int i;for(i=1;i<=n;i++)    if(father[i]==p)        {        dfs(i);        dp[p][1]+=dp[i][0];        dp[p][0]+=max(dp[i][0],dp[i][1]);        }}int main(){int i,t,a,b;while(~scanf("%d",&n))    {    memset(dp,0,sizeof(dp));    memset(father,0,sizeof(father));    for(i=1;i<=n;i++)        scanf("%d",&dp[i][1]); //如果当前选择的人来的话，最大欢乐值就是该人本身    while(scanf("%d%d",&a,&b)&&a+b)        father[a]=b;    i=n;    while(father[i]) //找出根节点，最大的上司        i=father[i];    dfs(i);    printf("%d\n",max(dp[i][0],dp[i][1]));    }return 0;}