integer input () pan rom list accep mes imm

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9354		Accepted: 5400

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

Source

Ural State University Internal Contest October‘2000 Students Session 題目大意：要開派對 n個人來派對會貢獻一定的快樂值 有n-1個關系 K L為L為K的直接上司 i和他的直接上司不能一起參加派對 求最大的快樂值 題解 樹形dp dp[i][0]為第i個點不去的最大值，dp[i][1]為第i個點去的最大值。 最後答案為dp[rt][0]和dp[rt][1]的最大值。 代碼

#include<iostream>
#include<cstdio>
using namespace std;

int n,x,y,rt;
int dp[6005][3],dad[6005];

void tree_dp(int now){
    for(int i=1;i<=n;i++)
     if(dad[i]==now){
         tree_dp(i);
         dp[now][1]+=dp[i][0];
         dp[now][0]+=max(dp[i][0],dp[i][1]);
     }
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&dp[i][1]);
    for(int i=1;i<n;i++)scanf("%d%d",&x,&y),dad[x]=y;
    rt=n;
    while(dad[rt])rt=dad[rt];
    tree_dp(rt);
    printf("%d",max(dp[rt][1],dp[rt][0]));
    return 0;
}

Anniversary party