1039. Anniversary Party
Time limit: 0.5 secondMemory limit: 8 MB
Background
The president of the Ural State University is going to make an 80’th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from 1 to N, The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.
Problem
Your task is to make up a guest list with the maximal conviviality rating of the guests.
Input
The first line of the input contains a number N. 1 ≤ N ≤ 6000.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 the supervisor relation tree goes.Each line of the tree specification has the form
<L> <K>
which means that the K-th employee is an immediate supervisor ofL-th employee. Input is ended with the line
0 0
Output
The output should contain the maximal total rating of the guests.
Sample
inputoutput
711111111 32 36 47 44 53 50 05
没有上司的年会。ural要举办场年会，请你邀请来宾。要求每位客人的直接上司不被邀请，否则会很尴尬。每位客人有一个兴趣值。要求使来宾的兴趣值之和最大。
每个人可以选择去或者不去。所以：
dp[t][0] 表示以t为最高上司的子树不安排t参加的最大兴趣值。
dp[t][1] 表示以t为最高上司的子树安排t参加的最大兴趣值。
不安排t参加的话，其直系下属可以去或者不去。dp[t][0] = sum(max(dp[ti][0], dp[ti][1]));
安排t参加的话，其直系下属均不能参加。dp[t][1] = sum(dp[ti][0]) + c[t];
#include <cstdio>#include <vector>#include <cstring>#include <iostream>using namespace std;int n;int c[6006];std::vector<int > v[6006];int dp[6006][2];void Tdp(int t) {if(v[t].size() == 0) {dp[t][0] = 0;dp[t][1] = c[t];return ;}for (int i=0; i<v[t].size(); i++) {Tdp(v[t][i]);}for (int i=0; i<v[t].size(); i++) {dp[t][0] += max(dp[v[t][i]][0], dp[v[t][i]][1]);dp[t][1] += dp[v[t][i]][0];}dp[t][1] += c[t];}int main () {cin >> n;for (int i=1; i<=n; i++) {cin >> c[i] ;}int l ,k;int vis[6006];memset(vis, false, sizeof(vis));while (cin >> l >> k) {if (l == 0 && k == 0) break;v[k].push_back(l);vis[l] = true;}int root;for (int i=1; i<=n; i++) {if (!vis[i]) {root = i;break;}}Tdp(root);cout << max(dp[root][0], dp[root][1]) <<endl;return 0;}
，教育人的诗句或名言警句,激励人在逆境中脱颖而出的话
