HDU 1520 & POJ 2342 Anniversary party(树形DP入门题)

题目链接

HDU: http://acm.hdu.edu.cn/showproblem.php?pid=1520

POJ: http://poj.org/problem?id=2342

Problem 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 T 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

 

Source

Ural State University Internal Contest October'2000 Students Session

题意：

一个聚会，需要邀请一些人来参加聚会来增加活跃度，但这N个人中，除了一个人以外，其余的人都有直接的上司，如果他们碰到他们的直接的上司的话，那么他们会很不愉快。

现在要求在所有的人都愉快的情况下，使得聚会的活跃度达到最大。

PS:

N个节点形成一棵树，根节点需要用到子节点的数据。

思路：先把每个点的直接的子节点找出来，存在邻接表里。再找出根节点，最后 dfs 从根节点出发找到子节点，然后数据再一层一层的返回。

dp[i][0] ：表示 i 这个节点不选， dp[i][1] 代表 i 这个节点选择了。
dp[i][0] += max (dp[子节点][0], dp[子节点][1]);上层节点不选了，那么子节点可选可不选。
dp[i][1] += dp[子节点][0]； 上层节点选了，那么子节点是不可选的。

代码如下：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define maxn 6010int dp[maxn][2];//dp[i][0]表示：不选 i，dp[i][1]:表示选择 iint vis[maxn];int a[maxn];int n;struct node{    int to;    int next;} e[maxn];int MAX(int a, int b){    if(a > b)        return a;    return b;}int cnt, head[maxn];void addEdge(int u, int v){    e[cnt].to = v;    e[cnt].next = head[u];    head[u] = cnt++;}void init(){    cnt = 0;    memset(head,-1,sizeof(head));    memset(dp,0,sizeof(dp));    memset(vis,0,sizeof(vis));}void dfs(int node){    dp[node][0] = 0;    dp[node][1] = a[node];    for(int i = head[node]; i != -1; i = e[i].next)    {        int u = e[i].to;        dfs(u);        dp[node][0] += MAX(dp[u][0], dp[u][1]);        dp[node][1] += dp[u][0];    }}int main(){    while(~scanf("%d",&n))    {        init();        for(int i = 1; i <= n; i++)        {            scanf("%d",&a[i]);        }        int root = 1;        int L, K;        while(scanf("%d%d",&L,&K))        {            if(L==0 && K==0)            {                break;            }            vis[L] = 1;            addEdge(K, L);        }        for(int i = 1; i <= n; i++)//寻找根节点        {            if(!vis[i])            {                root = i;                break;            }        }        dfs(root);        int ans = MAX(dp[root][0], dp[root][1]);        printf("%d\n",ans);    }    return 0;}/*711111111 32 36 47 44 53 50 0*/