Anniversary party(树形DP入门)
Anniversary party
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.
InputEmployees 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 0OutputOutput should contain the maximal sum of guests' ratings.
Sample Input
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0
Sample Output
5
题意：多组输入，每组输入的第一行一个N代表N个人，接下来N行每行一个数代表如果这个人参加party的欢乐值，接下来许多行，每行两个数u,v 表示v是u的直接上属。举行一个party，如果邀请了这个人的直接上司就不能邀请这个人，要求最后总的欢乐值最大，问欢乐值最大为多少。
思路：
开一个二维dp dp[x][0]代表没有邀请x的最大欢乐值、
dp[x][1]代表邀请了x的最大欢乐值。初始化dp[x][0]均为0 dp[x][1]为a[x]。dfs从树根遍历这棵树。具体看代码
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<algorithm>
6 using namespace std;
7 const int maxn=6010;
8 vector<int>v[maxn];
9 int dp[maxn][2],in[maxn],x,y;
10 int n,vis[maxn];
11 int a[maxn];
12 void dfs(int t)
13 {
14 dp[t][0]=0; //dp数组的初始化。
15 dp[t][1]=a[t];
16 vis[t]=1; //vis数组的意义如同记忆化搜索，如果这个叶子节点已经同时是其他节点的叶子节点，那么它已经被访问过，dp值已经更新了，不用再往下搜，因为下面都已经被搜过了，直接拿来用就行。
17 for(int i=0;i<v[t].size();i++)
18 {
19 int to=v[t][i];
20 if(vis[to]) //这个题数据比较水，如果不加vis数组也能过，但是时间会长。
21 continue;
22 dfs(to);
23 dp[t][0]+=max(dp[to][0],dp[to][1]);//如果不邀请t，那么就可以邀请他的直系下属
24 dp[t][1]+=dp[to][0]; //如果邀请t，那么久不能邀请他的直系下属
25 }
26 return ;
27 }
28 int main()
29 {
30 while(~scanf("%d",&n))
31 {
32 for(int i=1;i<=n;i++)
33 v[i].clear();
34 memset(in,0,sizeof(in));
35 memset(dp,0,sizeof(dp));
36 for(int i=1;i<=n;i++)
37 {
38 scanf("%d",&a[i]);
39 }
40 while(~scanf("%d%d",&x,&y)&&(x+y))
41 {
42 v[y].push_back(x);
43 in[x]++;
44 }
45 memset(vis,0,sizeof(vis));
46 for(int i=1;i<=n;i++)
47 {
48 if(!in[i])
49 {
50 dfs(i);
51 printf("%d\n",max(dp[i][0],dp[i][1]));
52 break;
53 }
54 }
55 }
56 }
