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07/25
22:09
DP 数据结构 算法区

HDU 4169 Wealthy Family

地址:戳这里

Problem Description

While studying the history of royal families, you want to know how wealthy each family is. While you have various ‘net worth’ figures for each individual throughout history, this is complicated by double counting caused by inheritance. One way to estimate the wealth of a family is to sum up the net worth of a set of k people such that no one in the set is an ancestor of another in the set. The wealth of the family is the maximum sum achievable over all such sets of k people.

Since historical records contain only the net worth of male family members, the family tree is a simple tree in which every male has exactly one father and a non-negative number of sons. You may assume that there is one person who is an ancestor of all other family members.

Input
The input consists of a number of cases. Each case starts with a line containing two integers separated by a space: N (1 <= N <= 150,000), the number of people in the family, and k (1 <= k <= 300), the size of the set. The next N lines contain two non-negative integers separated by a space: the parent number and the net worth of person i (1 <= i <= N). Each person is identified by a number between 1 and N, inclusive. There is exactly one person who has no parent in the historical records, and this will be indicated with a parent number of 0. The net worths are given in millions and each family member has a net worth of at least 1 million and at most 1 billion.
Output
For each case, print the maximum sum (in millions) achievable over all sets of k people satisfying the constraints given above. If it is impossible to choose a set of k people without violating the constraints, print ‘impossible’ instead.
Sample Input
5 3 0 10 1 15 1 25 1 35 4 45 3 3 0 10 1 10 2 10
Sample Output
85 impossible

w[i]表示以i为跟的子树中一个点的最大权值,f[i]表示在i为跟的子树下选择两个没有祖先关系的节点的最大权值和。一开始用堆放入一棵树中跟节点,优先级为w[i],每次选择堆顶的节点,统计答案增加w[i],然后把这个节点再次放入堆,优先级为f[i]-w[i],表示不选择i后这个子树选择多一个节点能增加的权值。如果堆首是已经取过的节点,从答案里面减少w[i],并且把i的所有儿子都放入堆里面。复杂度O(NlogN)。

代码:戳这里