poj2342——Anniversary party（树形dp）

最新推荐文章于 2020-05-14 10:04:03 发布

原创最新推荐文章于 2020-05-14 10:04:03 发布 · 603 阅读

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#poj #结构 #dp #acm #c

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本文介绍了一个基于树形结构的算法问题，目标是在遵循特定规则的情况下，最大化参与聚会员工的总欢乐值。输入包括员工间的层级关系及每个员工的欢乐值评分，通过递归深度优先搜索策略，动态规划求解最优解。

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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

先入个门

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <math.h>
#include <algorithm>
#include <queue>
#include <iomanip>
#define INF 0x3f3f3f3f
#define MAXN 10005
#define Mod 99999999
using namespace std;
int dp[MAXN][2],father[MAXN],vis[MAXN]; //dp[i][1]表示去,dp[i][0]表示不去
int n;
void dfs(int v)
{
    for(int i=1;i<=n;++i)
    {
        if(!vis[i]&&father[i]==v)
        {
            vis[i]=1;
            dfs(i);
            dp[v][1]+=dp[i][0];
            dp[v][0]+=max(dp[i][1],dp[i][0]);
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        memset(father,0,sizeof(father));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;++i)
            scanf("%d",&dp[i][1]);
        int l,k,root=1,flag=1;
        while(scanf("%d%d",&l,&k))
        {
            if(l==0&&k==0)
                break;
            father[l]=k;
            if(l==root||flag)
                root=l;
        }
        while(father[root])
            root=father[root];
        vis[root]=1;
        dfs(root);
        int ans=max(dp[root][1],dp[root][0]);
        printf("%d\n",ans);
    }
    return 0;
}