HDU - 1520 Anniversary party [树形dp]

最新推荐文章于 2024-05-22 17:32:19 发布

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最新推荐文章于 2024-05-22 17:32:19 发布

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原文链接：http://www.cnblogs.com/cniwoq/p/7246307.html

针对某大学举办的庆祝派对，每位员工拥有不同的活跃度评分，为了避免员工与直接上司同时出席，采用树形DP算法确定参与派对人员组合，以最大化整体活跃度。

Anniversary party

时限：1000ms

Problem 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 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 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

题意：

某某大学要开party，每个员工有一个活跃度。每个员工都不想跟他的直接上司见面，求party的最大活跃度。

思路：

树形dp，容易得到状态转移方程

dp[i][1] += dp[j][0]; i到下属j不到。

dp[i][0] +=max(dp[j][1],dp[j][0]); i不到。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 6010;
int n, dp[maxn][2], fa[maxn];  
vector<int> G[maxn];  
void dfs(int root) {
    for(int i=0;i<G[root].size();i++) {
        dfs(G[root][i]);  
    }  
    for(int i=0;i<G[root].size();i++) {
        dp[root][0] += max(dp[G[root][i]][0],dp[G[root][i]][1]);  
        dp[root][1] += dp[G[root][i]][0];  
    }  
}  
int main() {
    int u,v;  
    while(scanf("%d", &n) != EOF) {
        memset(fa,-1, sizeof(fa)); 
        memset(dp,0,sizeof(dp)); 
        for (int i = 1; i <= n; i++) { 
            scanf("%d", &dp[i][1]); G[i].clear();
        }  
        while(scanf("%d%d", &u,&v) != EOF) {
            if (v == 0 && v == 0) break;
            fa[u] = v;  
            G[v].push_back(u);  
        }  
        int root=1;  
        while(fa[root]!=-1) root=fa[root];  
        dfs(root);  
        printf("%d\n",max(dp[root][1],dp[root][0]));  
    }  
    return 0;  
}