題目連結

Anniversary Party

題目背景

The president of the Ural State University is going to make an 80'th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from \(1\)

to \(N\), The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.

題目描述

Your task is to make up a guest list with the maximal conviviality rating of the guests.

輸入格式

The first line of the input contains a number \(N\). \(1 \le N \le 6000\). 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 the supervisor relation tree goes. Each line of the tree specification has the form
\(<L>\ <K>\)

which means that the \(K\)-th employee is an immediate supervisor of \(L\)-th employee. Input is ended with the line
\(0\ 0\)

輸出格式

The output should contain the maximal total rating of the guests.

樣例輸入

7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0

樣例輸出

5

題解

題意：有一棵樹，每個點有一個權值，可以選擇一些點，要求一個點和它父親不能同時被選擇，求選擇的點的權值之和最大為多少。
很裸的樹上dp，我們令\(dp[i][1/0]\)表示\(i\)這個點選或不選，以\(i\)為根的子樹內權值之和最大為多少。
轉移方程就是：
\(dp[i][1]=\sum dp[j][0]\)
\(dp[i][0]=\sum max\)(\(dp[j][0],dp[j][1]\))
\(j\)為\(i\)的兒子。
上程式碼：

#include<bits/stdc++.h>
using namespace std;
int n;
int x,y;
int a[6009];
int dp[6009][2];
struct aa{
    int to,nxt;
}p[20009];
int h[6009],len=1;
void add(int u,int v){
    p[++len].to=v;
    p[len].nxt=h[u];
    h[u]=len;
}
void dfs(int u,int fa){
    int sum=0,ss=0;
    dp[u][0]=0;
    dp[u][1]=a[u];
    for(int j=h[u];j;j=p[j].nxt){
        if(p[j].to!=fa){
            dfs(p[j].to,u);
            dp[u][0]+=max(dp[p[j].to][0],dp[p[j].to][1]);
            dp[u][1]+=dp[p[j].to][0];
        }
    }
}
int main(){
    scanf("%d",&n);
    for(int j=1;j<=n;j++)
        scanf("%d",&a[j]);
    while(1){
        scanf("%d%d",&x,&y);
        if(x==0 && y==0) break;
        add(x,y);
        add(y,x);
    }
    dfs(1,0);
    printf("%d",max(dp[1][0],dp[1][1]));
    return 0;
}