HDU1520-Anniversary party

Anniversary party

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                    Total Submission(s): 10016    Accepted Submission(s): 4248

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

 

Source

Ural State University Internal Contest October'2000 Students Session

 

题意：有N个人，N-1个人有自己的上司，每个人有一个快乐值，如果这个人参加了聚会，那么这个人的直接上司将不能参加，问最大的快乐值为多少

题解：树形DP题，dp[i][0]表示第i个人不去，dp[i][1]表示第i个人去。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <set>using namespace std;int n;int f[10000],a[10000],dp[10000][2];int s[10000],e[10000],nt[10000];void dfs(int x){    if(s[x]==-1) return ;    dp[x][1]+=a[x];    for(int i=s[x];i!=-1;i=nt[i])    {        int ee=e[i];        dfs(ee);        dp[x][1]+=dp[ee][0];        dp[x][0]+=max(dp[ee][1],dp[ee][0]);    }}int main(){    while(~scanf("%d",&n))    {        memset(f,0,sizeof f);        memset(s,-1,sizeof s);        memset(nt,-1,sizeof nt);        memset(dp,0,sizeof dp);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        int u,v,sum=0;        while(1)        {            scanf("%d %d",&u,&v);            if(!u&&!v) break;            int flag=1;            for(int i=s[v];i!=-1;i=nt[i])                if(e[i]==u) {flag=0;break;}            if(!flag) continue;            nt[++sum]=s[v];            s[v]=sum;            e[sum]=u;            f[u]++;        }        int k;        for(int i=1;i<=n;i++)        {            if(!f[i]) k=i;            if(s[i]==-1) dp[i][1]=a[i];        }        dfs(k);        printf("%d\n",max(dp[k][0],dp[k][1]));    }    return 0;}