A - Anniversary party HDU - 1520（没有上司的舞会）-白红宇

A - Anniversary party HDU - 1520（没有上司的舞会）

阅读量：4136 次

发布时间：2019-05-25

本文共 2458 字，大约阅读时间需要 8 分钟。

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

树形dp入门题目，上司和下属不能同时出现。。就要动态规划下去，看看怎样的布置才会使总价值最高。我觉得树形dp的题目还是需要用vector，虽然用数组啥的在poj上可以过去，但是在hdu上过不了。卡数据卡的太厉害了。

代码如下：

//正确代码：#include
   
    #include
    
     #include
     
      using namespace std;const int maxn=6005,inf=1<<30;int dp[maxn][2];int w[maxn],in[maxn];int n,m;vector
      
       G[maxn];void dfs(int u,int fa){     dp[u][1]=w[u];    dp[u][0]=0;    for(int i=0;i

但是我还是觉得把那个代码贴出来，毕竟在poj上可以过去哦

代码如下：

（错误代码）在hdu上过不去的，呜呜呜

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        using namespace std;const int maxx=6e3+10;int dp[maxx][2];int vis[maxx];int ins[maxx];int f[maxx];void init(){	memset(dp,0,sizeof(dp));	memset(vis,0,sizeof(vis));	memset(ins,0,sizeof(ins));	memset(f,0,sizeof(f));}int n,l,k,top;void dfs(int top){	vis[top]=1;	for(int i=1;i<=n;i++)	{		if(vis[i]==0&&f[i]==top)		{			dfs(i);			dp[top][1]+=dp[i][0];			dp[top][0]+=max(dp[i][0],dp[i][1]);		}	}}int main(){	while(scanf("%d",&n)!=EOF)	{		for(int i=1;i<=n;i++) cin>>dp[i][1];		while(scanf("%d%d",&l,&k),l+k)		{			f[l]=k;			ins[l]++;		}		for(int i=1;i<=n;i++)		{			if(ins[i]==0)			{				top=i;				break;			}		}		dfs(top);		cout<
        
         <

努力加油a啊，(^o)/~

转载地址：http://rzxvi.baihongyu.com/

你可能感兴趣的文章