作者：Gome--李想 | 来源：互联网 | 2023-09-25 12:13
本文由编程笔记#小编为大家整理，主要介绍了HDU-1520Anniversaryparty(有向入门树形DP)相关的知识，希望对你有一定的参考价值。题目链接：http
本文由编程笔记#小编为大家整理，主要介绍了HDU - 1520 Anniversary party (有向入门树形DP)相关的知识，希望对你有一定的参考价值。
题目链接：https://vjudge.net/problem/HDU-1520
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.
InputEmployees 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 0OutputOutput 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
题意：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：从根出发找，dp[i][0]表示节点i不选，dp[i][1]节点i选。
AC代码：
#include
#include
#include
#include
using namespace std;
const int maxn=6010;
vectormap[maxn];
int dp[maxn][2],a[maxn],N,vis[maxn];
void dfs(int u){
dp[u][0]=0;
dp[u][1]=a[u];
for(int i=0;i {
int v=map[u][i];
dfs(v);
dp[u][0]+=max(dp[v][0],dp[v][1]);
dp[u][1]+=dp[v][0];
}
}
int main(){
while(~scanf("%d",&N)){
for(int i=1;i<=N;i++) scanf("%d",&a[i]);
for(int i=1;i<=N;i++) map[i].clear();
memset(vis,0,sizeof(vis));
int u,v;
while(scanf("%d%d",&v,&u)&&u&&v)
{
map[u].push_back(v);
vis[v]++;
}
for(int i=1;i<=N;i++)
if(!vis[i]){
dfs(i);
printf("%d
",max(dp[i][0],dp[i][1]));
break;
}
}
return 0;
}
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