小白算法练习 树状dp POJ anniversary party

Anniversary party

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9493 Accepted: 5464

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 N – 1 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

#include<iostream>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int Max=6008;vector<int>vec[Max];int N;int value[Max];int vis[Max];int root[Max];int dp[Max][2]; void init(){for(int i=1;i<=N;i++){scanf("%d",&value[i]);}int f,e;while(scanf("%d%d",&f,&e)!=EOF && f!=0 && e!=0){vec[e].push_back(f);vis[f]=1; //如果是下属则置1}for(int i=1;i<=N;i++)//找根{if(vis[i]!=1){vec[0].push_back(i);//根节点 有可能有多个根 比如有多个经理 }}}void dfs(int rt){dp[rt][0]=0;dp[rt][1]=value[rt];for(int i=0;i<vec[rt].size();i++){dfs(vec[rt][i]);dp[rt][0]+=max(dp[vec[rt][i]][0],dp[vec[rt][i]][1]);dp[rt][1]+=dp[vec[rt][i]][0];}}void solve(){int ans=0;for(int i=0;i<vec[0].size();i++) //把所有根得到的结果加起来{dfs(vec[0][i]);ans+=max(dp[vec[0][i]][0],dp[vec[0][i]][1]);}printf("%d\n",ans);}int main(){ while(scanf("%d", &N) != EOF) {for(int i=0;i<Max;i++) vec[i].clear();memset(value,0,sizeof(value));memset(vis,0,sizeof(vis));memset(dp,0,sizeof(dp));        init();                solve();            }    return 0;}