HDU1520 Anniversary party（树形dp）

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8046    Accepted Submission(s): 3507

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：举办一个派对，想邀请校内职员参加，但是不同时邀请员工的该员工的直属上级，每位职员都有一个评级，问怎样邀请能使评级之和最大？

思路：刚开始用dp[i][j][0 or 1]表示第i个职员包括他的下属一共邀请j个能获得的最大值，dp[i][j][0]表示第i个不邀请，dp[i][j][1]表示邀请第i个。后来内存超了。

现在想来为什么一定要限制人数......所以应该用dp[i][0 or 1]这样就可以了，dp[fa][0] += max(dp[son][0], dp[son][1]), dp[fa][1] += dp[son][0]。这样就是普通的树形dp了。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define maxn 6005int dp[maxn][2], val[maxn], head[maxn], tot, N;bool vis[maxn];struct Edge{    int to, next;}edge[maxn*2];void add(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void dfs(int fa){    int i, j, k, son;    vis[fa] = 1;    dp[fa][1] = val[fa];    for(i = head[fa];i != -1;i = edge[i].next){        son = edge[i].to;        //printf("fa:%d   son:%d\n", fa, son);        if(vis[son]) continue;        dfs(son);        dp[fa][0] += max(dp[son][0], dp[son][1]);        dp[fa][1] += dp[son][0];    }}int main(){    int i, u, v, rt;    while(scanf("%d", &N) != EOF){        tot = 1;        memset(head, -1, sizeof head);        memset(vis, 0, sizeof vis);        memset(dp, 0, sizeof dp);        for(i = 1;i <= N;i++) scanf("%d", &val[i]);        while(scanf("%d %d", &v, &u)){            if(!u&&!v) break;            add(u, v);            vis[v] = true;        }        for(i = 1;i <= N;i++){            if(!vis[i]) rt = i;            vis[i] = false;        }        //printf("%d\n", rt);        dfs(rt);        printf("%d\n", max(dp[rt][0], dp[rt][1]));    }}