【hdu1520-Anniversary party】树形DP

最新推荐文章于 2024-05-22 17:32:19 发布

原创最新推荐文章于 2024-05-22 17:32:19 发布 · 199 阅读

0 ·

CC 4.0 BY-SA版权

动态规划-DP 专栏收录该内容

24 篇文章

订阅专栏

针对大学80周年庆典，通过员工间的层级关系及其亲和力评分，设计算法选出总评分最高的嘉宾组合，确保直接上下级不会同时出席。

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

1 3

2 3

6 4

7 4

4 5

3 5

0 0

Sample Output

5

【大意】

在一个有根树上每个节点有一个权值，每相邻的父亲和孩子只能选择一个，问怎么选择总权值之和最大。

【思路】

dp[i][0/1]代表当前节点不选/选能得到的最大权值， max(dp[root][0], dp[root][1])即所求

转移方程dp[u][1] += dp[v][0], dp[u][0] += max(dp[v][0], dp[v][1])

初始化dp[u][0] = 0, dp[u][1] = happy[u]

【代码】

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 6003;
int dp[N][2], happy[N], in[N];
vector<int>G[N];
void dfs(int u)
{
    dp[u][0] = 0, dp[u][1] = happy[u];
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        dfs(v);
        dp[u][0] += max(dp[v][0], dp[v][1]);
        dp[u][1] += dp[v][0];
    }
}
int main()
{
    int n, i, a, b;
    while(~scanf("%d", &n))
    {
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &happy[i]);
            G[i].clear();
            in[i] = 0;
        }
        memset(dp, 0, sizeof(dp));
        while(scanf("%d%d", &a, &b), a+b)
        {
            G[b].push_back(a);//这里加边顺序不能错
            in[a]++;
        }
        for(i = 1; i <= n; i++) if(!in[i]) break;//入度为0即为根
        dfs(i);
        printf("%d\n", max(dp[i][0], dp[i][1]));
    }
    return 0;
}