POJ2342 Anniversary party

Anniversary party
 

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 N – 1 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

5

树型DP.

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<cstdlib>
 7 #include<vector>
 8 using namespace std;
 9 typedef long long ll;
10 typedef long double ld;
11 typedef pair<int,int> pr;
12 #define rep(i,a,n) for(int i=a;i<=n;i++)
13 #define per(i,n,a) for(int i=n;i>=a;i--)
14 #define Rep(i,u) for(int i=head[u];i;i=Next[i])
15 #define clr(a) memset(a,0,sizeof(a))
16 #define pb push_back
17 #define mp make_pair
18 #define fi first
19 #define sc second
20 #define pq priority_queue
21 #define pqb priority_queue <int, vector<int>, less<int> >
22 #define pqs priority_queue <int, vector<int>, greater<int> >
23 #define vec vector
24 ld eps=1e-9;
25 ll pp=1000000007;
26 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;}
27 ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;}
28 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
29 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; }
30 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1};
31 ll read(){ ll ans=0; char last=' ',ch=getchar();
32 while(ch<'0' || ch>'9')last=ch,ch=getchar();
33 while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
34 if(last=='-')ans=-ans; return ans;
35 }
36 const int N=10000;
37 int v[N],Next[N],head[N],dp[N][2],a[N],in[N],e;
38 void add(int x,int y){ v[++e]=y; Next[e]=head[x]; head[x]=e; }
39 void dfs(int x,int f){
40     for (int i=head[x];i;i=Next[i])
41     if (v[i]){
42         dfs(v[i],x);
43         dp[x][1]+=dp[v[i]][0];
44         dp[x][0]+=max(dp[v[i]][0],dp[v[i]][1]);
45     }
46 }
47 int main(){
48     int n=read();
49     for (int i=1;i<=n;i++) a[i]=read(),dp[i][1]=a[i];
50     int a_,b_,root;
51     while (~scanf("%d%d",&a_,&b_)){
52         if (a_==0 && b_==0) break;
53         add(b_,a_); in[a_]++;
54     }
55     for (int i=1;i<=n;i++)
56         if (!in[i]) root=i;
57     dfs(root,-1);
58     printf("%d",max(dp[root][1],dp[root][0]));
59     return 0;
60 }
View Code

 

posted @ 2017-10-08 21:12  SXia  阅读(146)  评论(0)  编辑  收藏  举报