poj 2342 Anniversary party 树形DP 解题报告

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

题意

话说一个公司的一些然要去参加一个party，每个人有一个愉悦值，而如果某个人的直接上司在场的话会非常扫兴，所以避免这样的安排，问给出n个人，每个人的愉悦值以及他们的上司所属关系，问你让那些人去可以让总的愉悦值最大，并求出这个值。

思路

树形dp入门题目，这个公司的人事关系可以根据给出的数据得到一个树，最上面的是最高层，往下依次，我们要做的就是在树的节点处进行dp。
用dp【i】【0】表示当前i这个人不选，dp【i】【1】表示当前i这个人安排去参加。
那么dp【st】【1】+=dp【i】【0】 ///因为当前这个要选，那么他的前一个一定不能选，否则不满足题目要求
而 dp【st】【0】+=max（dp【i】【0】，dp【i】【1】） 而在树中 st 是 i 的父节点。
实现的话很简单，就是直接找到根节点，建树的过程中dp，最后得到结果

代码

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<vector>using namespace std;const int N=10000;  int n,dp[N][3],father[N],flag[N];//dp[i][0]表示当前i点不选 1表示选  void creat(int x)  {      flag[x]=1;      for(int i=1;i<=n;i++)      if(flag[i]==0&&father[i]==x)      {          creat(i);          dp[x][0]+=max(dp[i][0],dp[i][1]);          dp[x][1]+=dp[i][0];      }  }  int main()  {      while(~scanf("%d",&n))      {          memset(dp,0,sizeof(dp));        memset(father,0,sizeof(father));        memset(flag,0,sizeof(flag));        for(int i=1;i<=n;i++)          scanf("%d",&dp[i][1]);          int f,c,root;           root=0;//记录父结点          int beg=1;          while (scanf("%d %d",&c,&f),c||f)          {              father[c]=f;              if(root==c||beg) root=f;            }          while(father[root])//查找父结点          root=father[root];          creat(root);          int imax=max(dp[root][0],dp[root][1]);          printf("%d\n",imax);      }      return 0;  } 

当然，还有升级版

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>#include<vector>using namespace std;  const int N=6000+5;  int val[N],n,dp[N][2],vis[N];  vector<int> vec[N];  void dfs(int i,int pre){    vis[i]=true;    int sum=0,tot=0;    for (int j=0;j<vec[i].size();j++)    {        int temp=vec[i][j];        if (vis[temp]) continue;        dfs(temp,i);        sum+=max(dp[temp][1],dp[temp][0]);        tot+=dp[temp][0];    }    dp[i][0]=sum;    dp[i][1]=val[i]+tot;}int main(){    vec[0].push_back(1);    while(scanf("%d",&n)!=EOF)    {        memset(vis,0,sizeof(vis));        memset(dp,0,sizeof(dp));        for (int i=1;i<=n;i++)        {            scanf("%d",&val[i]);            vec[i].clear();        }        while(true)        {            int a,b;            scanf("%d%d",&a,&b);            if (!a) break;            vec[a].push_back(b);            vec[b].push_back(a);        }        dfs(0,0);        printf("%d\n",dp[0][0]);    }    return 0;}