Anniversary party
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12770 Accepted Submission(s): 5142
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
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
5
Source
Recommend
题意：有n个人，他们之间有上下级关系并且关系形成了一棵树，现在要求直接上下级的人不能同时参加聚会，每个人参加都有一个快乐值，问最大的快乐值。
思路：基础的树形dp题，dp[i][0]表示i不参加聚会时最大的快乐值；dp[i][1]表示i参加聚会时最大的快乐值。那么就有如下关系：
dp[u][1]+=dp[v][0];当u参加市，u的儿子v均不参加
dp[u][0]+=max(dp[v][1],dp[v][0]);当u不参加时，u的儿子可以参加，也可以不参加，取最大。
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<bitset>
using namespace std;
#define PI acos(-1.0)
#define eps 1e-8
typedef long long ll;
typedef pair<int,int> P;
const int N=1e5+,M=1e5+;
const int inf=0x3f3f3f3f;
const ll INF=1e18+,mod=1e9+;
struct edge
{
int from,to;
int next;
};
edge es[M];
int cnt,head[N];
int in[N];
int dp[N][];
void init()
{
cnt=;
memset(head,-,sizeof(head));
memset(in,,sizeof(in));
memset(dp,,sizeof(dp));
}
void addedge(int u,int v)
{
cnt++;
es[cnt].from=u,es[cnt].to=v;
es[cnt].next=head[u];
head[u]=cnt;
}
void dfs(int u)
{
for(int i=head[u]; i!=-; i=es[i].next)
{
int v=es[i].to;
dfs(v);
dp[u][]+=dp[v][];
dp[u][]+=max(dp[v][],dp[v][]);
}
///cout<<u<<" * "<<dp[u][0]<<" * "<<dp[u][1]<<endl;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
init();
for(int i=; i<=n; i++) scanf("%d",&dp[i][]);
int u,v;
while(scanf("%d%d",&u,&v)&&!(u==&&v==))
{
addedge(v,u);
in[u]++;
}
int root;
for(int i=; i<=n; i++)
if(!in[i]) root=i;
dfs(root);
printf("%d\n",max(dp[root][],dp[root][]));
}
return ;
}
基础树形dp