Hdu 1520 树形DP 解题报告

最新推荐文章于 2021-05-27 19:49:04 发布

原创最新推荐文章于 2021-05-27 19:49:04 发布 · 457 阅读

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#hdu #树形DP

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为庆祝大学80周年，需组织一场员工聚会。由于员工间的层级关系，不能同时邀请员工及其直接上司。通过树形动态规划算法，确定最佳嘉宾名单以最大化聚会氛围指数。

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Anniversary party

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11018 Accepted Submission(s): 4567

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.

题解

树形dp的常规入门题：设dp[i][0]表示:当前这个点不选，dp[i][1]表示当前这个点选择的最优解。

转移方程：

dp[cur][0]+=max(dp[son][1],dp[son][0]);//当前这个点不选，那他的孩子可选可不选，取
最大的。
dp[cur][1]+=dp[son][0]//当前这点选择，那他的孩子就不能选择。

代码：

AC代码如下

#include<iostream>  
#include<string>  
#include<cstring>  
#include<algorithm>  
#include<vector>  
#include<cstdio>
using namespace std;  
const int N=6001;  
int val[N],n,dp[N][2];  
vector<int> vec[N];  
bool vis[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)  
    {  
        for (int i=1;i<=n;i++) 
        {
            scanf("%d",val+i);
            vec[i].clear();
        }
        int a,b;  
        while (true)  
        {  
            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]);  
        memset(vis,false,sizeof(vis));  
        memset(dp,0,sizeof(dp));      
    }
    return 0;  
}

还有一个爆内存的程序，不知道为什么

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<cmath>
using namespace std;  
const int N=6000+10;
int pre[N],dp[N][2],vis[N],n;
vector<int> G[N];
int max(int x,int y)
{
    if (x>=y) return x;
    else return y;
}
void dfs(int u,int father)
{
    int d=G[u].size();
    for (int i=0;i<d;i++)
    {
        int v=G[u][i];
        if (v!=father) dfs(v,pre[v]=u);
    }
}  
void find(int t)
{
    vis[t]=1;
    for (int i=1;i<=n;i++)
    {
        if (pre[i]==t&&!vis[i])
        {
            find(i);
            dp[t][0]+=max(dp[i][0],dp[i][1]);
            dp[t][1]+=dp[i][0];
        }
    }
}
int main()
{
    int u,v;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for (int i=1;i<=n;i++)
        scanf("%d",&dp[i][1]);
        while(scanf("%d%d",&u,&v)&&(u+v))
        {
            G[u].push_back(v);
            G[v].push_back(u);
        }
        memset(pre,-1,sizeof(pre));
        dfs(1,-1);
        memset(vis,0,sizeof(vis));
        find(1);
        printf("%d\n",max(dp[1][0],dp[1][1])); 
    }
    return 0;
}