hdoj 1520 Anniversary party 【树形DP入门】

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5954    Accepted Submission(s): 2705

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
   
   
   
   

第一道树形dp： 

 

#include<stdio.h>
#include<string.h>
#include<vector>
#define MAX 6000+10
using namespace std;
int max(int x,int y)
{
    return x>y?x:y; 
}
int dp[MAX][2],vis[MAX];
vector<int>tree[MAX];//存储子节点 
void dfs(int node)
{
    vis[node]=1;
    int i,j;
    int son;//子节点 
    for(i=0;i<tree[node].size();i++)
    {
        son=tree[node][i];
        if(!vis[son])
        {
            dfs(son);
            dp[node][1]+=dp[son][0];//上司去 下属不去 
            dp[node][0]+=max(dp[son][1],dp[son][0]); // 上司不去 下属去或不去 
        } 
    }
}
int main()
{
    int n,i;
    int root;//根节点 
    int x,y;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&dp[i][1]);
            dp[i][0]=0;
            vis[i]=1; 
            tree[i].clear();//清空 
        }
        while(scanf("%d%d",&x,&y)&&(x!=0||y!=0))
        {
            vis[x]=0;//x为下属 标记一下 
            tree[y].push_back(x);//建立关系 
        }
        root=0;
        for(i=1;i<=n;i++)//查询根节点 
        {
            if(vis[i])
            {
                root=i;//记录根节点 
                break;
            }
        }
        memset(vis,0,sizeof(vis));
        dfs(root);
        printf("%d\n",max(dp[root][0],dp[root][1]));
    }
    return 0;
} 

更新：2015.8.11

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#define MAXN 6000+10
using namespace std;
vector<int> G[MAXN];
int dp[MAXN][2];
int pre[MAXN];
int N;
int find(int p)
{
    int t;
    int child = p;
    while(child != p)
        p = pre[p];
    while(child != p)
    {
        t = pre[child];
        pre[child] = p;
        child = t;
    }
    return p;
}
void merge(int x, int y)
{
    int fx = find(x);
    int fy = find(y);
    if(fx != fy)
        pre[fx] = fy;
}
void DFS(int u)
{
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        DFS(v);
        dp[u][1] += dp[v][0];
        dp[u][0] += max(dp[v][1], dp[v][0]);
    }
}
int main()
{
    while(scanf("%d", &N) != EOF)
    {
        int a, b, root;
        for(int i = 1; i <= N; i++)
        {
            scanf("%d", &a);
            dp[i][1] = a;//i去
            dp[i][0] = 0;//i不去
            pre[i] = i;
            G[i].clear();
        }
        while(scanf("%d%d", &a, &b), a||b)
            G[b].push_back(a), pre[a] = b;
        for(int i = 1; i <= N; i++)
        {
            if(pre[i] == i)
                root = i;
        }
        DFS(root);
        printf("%d\n", max(dp[root][0], dp[root][1]));
    }
    return 0;
}