hdu 1520Anniversary party 树形dp入门

最新推荐文章于 2018-12-20 10:33:57 发布

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最新推荐文章于 2018-12-20 10:33:57 发布

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原文链接：http://www.cnblogs.com/qldabiaoge/p/9345703.html

本文介绍了一个用于选择参加大学80周年庆典的员工名单的算法。该算法通过递归求解，确保不会同时邀请到员工及其直接上司，并最大化受邀员工的亲和力评分总和。输入包括员工数量、亲和力评分及上下级关系。

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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

Sample Output

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn = 6e3 + 10;
 6 const int INF = 0x7fffffff;
 7 int n, dp[maxn][2], fa[maxn], vis[maxn];
 8 void solve(int x) {
 9     vis[x] = 1;
10     for (int i = 1 ; i <= n ; i++) {
11         if (!vis[i] && fa[i] == x ) {
12             solve(i);
13             dp[x][0] += max(dp[i][0], dp[i][1]);
14             dp[x][1] += dp[i][0];
15         }
16     }
17 }
18 int main() {
19     while(scanf("%d", &n) != EOF) {
20         memset(vis,0,sizeof(vis));
21         memset(dp,0,sizeof(dp));
22         memset(fa,0,sizeof(fa));
23         for (int i = 1 ; i <= n ; i++) scanf("%d", &dp[i][1]);
24         int x, y, root = 0;
25         while(scanf("%d%d", &x, &y)) {
26             if (x == 0 && y == 0) break;
27             fa[x] = y;
28             root = y;
29         }
30         while(fa[root]) root = fa[root];
31         solve(root);
32         printf("%d\n", max(dp[root][0], dp[root][1]));
33     }
34     return 0;
35 }