HDU1520 树形DP裸题

最新推荐文章于 2019-07-02 17:05:01 发布

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为庆祝大学80周年校庆，需邀请员工参加派对，并确保员工及其直接上司不会同时出席。通过评估每位员工的亲和力，采用动态规划算法找出邀请名单，使得总亲和力最大。

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
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

第一次做这种树形的DP
这个左孩子右兄弟的用法我还用的不是很熟练
导致WA一晚上。。

WA的主要原因在于我在建这个树的时候搞错了右边兄弟的地址储存
导致每多一个兄弟就会把之前的兄弟给覆盖掉
而我的测试用例却又是只有二叉树
根本找不到错误的样例….
这个错误要记住…

至于这个题本身只是一个弱智DP没什么亮点

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<memory.h>
using namespace std;
struct p
{
    long long quan,ba,er,nx,ex;
};
p zhi[6001];
long long dp[6002][2];
void build(long long l,long long k)
{
    zhi[l].ba=k;
    if(zhi[k].er==0)
    {
        zhi[k].er=l;
        zhi[k].ex=l;
    }
    else
    {
        zhi[zhi[k].ex].nx=l;
        zhi[k].ex=l;
    }
}
void NineFailure(long long root)
{
    if(zhi[root].er==0)
    {
        dp[root][0]=0;
        dp[root][1]=zhi[root].quan;
        return;
    }
    long long sum0=0;
    long long sum1=0;
    long long head=zhi[root].er;
    for(;zhi[head].ba==root;head=zhi[head].nx)
    {
        NineFailure(head);
        sum1+=dp[head][0];
        sum0+=max(dp[head][1],dp[head][0]);
    }
    dp[root][0]=sum0;
    dp[root][1]=sum1+zhi[root].quan;
}
int main()
{
    int n;
    while(cin>>n)
    {
        memset(zhi,0,sizeof(zhi));
        memset(dp,0,sizeof(dp));
        for(long long a=1;a<=n;a++)scanf("%lld",&zhi[a].quan);
        long long l,k;
        while(1)
        {
            scanf("%lld%lld",&l,&k);
            if(l+k==0)break;
            build(l,k);
        }
        long long sum;
        for(long long a=1;a<=n;a++)
        {
            if(zhi[a].ba==0)
            {
                NineFailure(a);
                sum=max(dp[a][1],dp[a][0]);
                break;
            }
        }
        cout<<sum<<endl;
    }
    return 0;
}