POJ 2342 Anniversary party（树形dp）

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7230		Accepted: 4162

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

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

Source

Ural State University Internal Contest October'2000 Students Session

 

思路

题意：某公司要举办一次晚会，但是为了使得晚会的气氛更加活跃，每个参加晚会的人都不希望见到自己的直接上司。现在已知每个人的活跃指数和上司关系（不存在环），求晚会的活跃指数的最大值。
思路：dp[x][0]表示x去参加晚会，dp[x][1]表示x不去参加晚会。那么有以下两种情况：

• x去参加晚会，则他的直接下属y就不能参加晚会，dp[ x ][ 1 ] = dp[ y [ 0 ]
• x不去参加晚会，则他的直接下属y可以参加也可以不去参加，dp[ x ][ 0 ] = max ( dp[ y ][ 0 ],dp[ y ][ 1 ])

 

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/*     dp[x][0]:x 不去参加聚会     dp[x][1]:x 去参加聚会 */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn =  6005; int tot = 0,fa[maxn],head[maxn],dp[maxn][2];; struct Tree{     int fa,son;     int next;     Tree():fa(0),son(0),next(0){} }tree[maxn];   void addedge(int u,int v) {     tree[tot].son = v;     tree[tot].next = head[u];     head[u] = tot++; }   void dfs(int cur) {     for (int i = head[cur];i != -1;i = tree[i].next)     {         dfs(tree[i].son);         dp[cur][1] += dp[tree[i].son][0];         dp[cur][0] += max(dp[tree[i].son][0],dp[tree[i].son][1]);     } }     int main() {     int N;     while (~scanf("%d",&N))     {         memset(head,-1,sizeof(head));         memset(dp,0,sizeof(dp));         for (int i = 1;i <= N;i++)   scanf("%d",&dp[i][1]);         int L,K;         while (scanf("%d%d",&L,&K) && L && K)         {             tree[L].fa = K;             addedge(K,L);         }         int root = 1;         while (tree[root].fa)   root = tree[root].fa;         dfs(root);         printf("%d\n",max(dp[root][0],dp[root][1]));     }     return 0; }