POJ 2342 Anniversary party（树形DP）

原题地址

Anniversary party

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7291 Accepted: 4195

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

711111111 32 36 47 44 53 50 0

Sample Output

5

树形DP，实际上就像是把入门级的数塔放在了树上。

记dp[ i ][ 0 ]为 i 不去的下属的最大活跃度和。

记dp[ i ][ 1 ]为 i 去的时候，i 及其下属的最大活跃度和。

若 j 为 i 的直接下属。

dp[ i ][ 1 ] = dp[ j ][ 0 ] //  如果 i 去了，她的直接下属 i 不能去。

dp[ i ][ 0 ] = max ( dp[ j ][ 1 ] ，dp[ j ][ 0 ] ) //如果 i 不去，那她的直接下属可以去或者不去。

我们从叶子节点开始，一直向根节点进行决策。

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int par[6066];bool vis[6066];int dp[6066][3];int n;void init(int n){    for(int i=1;i<=n;i++)    {        par[i]=i;//初始化i的父节点为i(我参考的并查集的写法)        vis[i]=0;//检查i是否已经进行过决策        dp[i][1]=0;        dp[i][0]=0;    }}void dfs(int root){    vis[root]=1;    for(int i=1;i<=n;i++)    {        if(!vis[i]&&par[i]==root)        {            dfs(i);//dfs要在前，更新DP要在后。因为我们是从叶子节点开始的。            dp[root][1]+=dp[i][0];            dp[root][0]+=max(dp[i][0],dp[i][1]);        }    }}int main(){    while(~scanf("%d",&n))    {        init(n);        int a,b;        for(int i=1;i<=n;i++)        {            scanf("%d",&dp[i][1]);        }        while(scanf("%d%d",&a,&b)&&(a||b))        {            par[a]=b;        }        int root=1;        while(par[root]!=root)        {            root=par[root];        }        dfs(root);        printf("%d\n",max(dp[root][1],dp[root][0]));    }    return 0;}

我们从叶子节点开始。