poj 2342 Anniversary party（树形DP基础题）（树形dp模板）

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5888 Accepted: 3389

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

Source

Ural State University Internal Contest October'2000 Students Session

题目大意：

       一家公司要举办年会，公司有n个人，每个人有价值vi，这家公司有严格的等级制度，参加活动时，不能同时出现a和a的直接上司，求最大价值总和

解题思路：

       下属关系属于一个树形（森林）结构，因为除了boss（们）每个人有一个上司，可以连一条边，dp[i][j]表示第i人参加（j==1）或者不参加（j==0）他和的他的子树所能获得的最大价值，那么dp[i][0]=sigema(max(dp[j][0],dp[j][1]))(j是子节点）,dp[i][1]=v[i]+sigema(dp[j][0])(j是子节点），要求的就是max(dp[1][0],dp[1][1]);

 p.s.注意有多个没有上司的boss根节点。

//使用向量124ms，链表做109ms，差的不多#include<stdio.h>#include<map>#include<string>#include<cstring>#include<algorithm>#include<vector>#include<iostream>#define maxn 6005#define c(a) memset(a,0,sizeof(a))#define c_1(a) memset(a,-1,sizeof(a))using namespace std;vector<int>cl[maxn];int  dp[maxn][2];short r[maxn];bool v[maxn];int dfs(int i, int j){if (dp[i][j] != -1)return dp[i][j];int ans = 0;if (j == 0){for (int k = 0; k < cl[i].size(); k++){int a = dfs(cl[i][k], 0);int b = dfs(cl[i][k], 1);ans += max(a, b);}return dp[i][j] = ans;}else{ans+=r[i];for (int k = 0; k < cl[i].size(); k++){ans += dfs(cl[i][k], 0);}return dp[i][j] = ans;}}int main(){int n;while (scanf("%d", &n)==1){c(r); c_1(dp); c(v);for (int i = 0; i<=n; i++)cl[i].clear();int a, b;for (int i = 1; i <=n; i++) scanf("%d", &r[i]);while(scanf("%d%d", &a, &b)==2&&a&&b){cl[b].push_back(a);v[a] = 1;}for (int i = 1; i <= n; i++)if (cl[i].size() == 0) { dp[i][0] = 0; dp[i][1] = r[i]; }int x=0;for (int i = 1; i <= n; i++)if (!v[i]) x = x+max(dfs(i, 1), dfs(i, 0)); printf("%d\n",x);}}