[HDU 1520] Anniversary party

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6078    Accepted Submission(s): 2752

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

 

Source

Ural State University Internal Contest October'2000 Students Session

 

树形DP入门题目、、

加了一个输入输出外挂、、无耻地飘进首页、、

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <queue>

#include <cmath>

#include <map>

#include <vector>

#include <iterator>

#include <cstring>

#include <string>

using namespace std;

#pragma comment(linker, "/STACK:1024000000,1024000000")

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define INF 0x7fffffff

#define ll long long

#define N 6010



struct Edge

{

    int to,next;

}edge[N<<1];

int head[N],tot;

int n;

int f[N];

int val[N];

int vis[N];

int dp[N][2];



template <class T>

inline bool input(T &ret)

{

    char c; int sgn;

    if(c=getchar(),c==EOF) return 0;

    while(c!='-'&&(c<'0'||c>'9')) c=getchar();

    sgn=(c=='-')?-1:1;

    ret=(c=='-')?0:(c-'0');

    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');

    ret*=sgn;

    return 1;

}



inline void out(int x)

{

    if(x>9) out(x/10);

    putchar(x%10+'0');

}



void init()

{

    tot=0;

    memset(head,-1,sizeof(head));

    memset(f,0,sizeof(f));

    memset(dp,0,sizeof(dp));

    memset(vis,0,sizeof(vis));

}



void add(int x,int y)

{

    edge[tot].to=y;

    edge[tot].next=head[x];

    head[x]=tot++;

}



void dfs(int u)

{

    for(int i=head[u];i!=-1;i=edge[i].next)

    {

        int v=edge[i].to;

        if(!vis[v])

        {

            vis[v]=1;

            dfs(v);

            dp[u][1]+=dp[v][0];

            dp[u][0]+=max(dp[v][0],dp[v][1]);

        }

    }

    dp[u][1]+=val[u];

}



int main()

{

    int i;

    while(input(n))

    {

        init();

        for(i=1;i<=n;i++)

        {

            input(val[i]);

        }

        int a,b;

        while(1)

        {

            input(a);

            input(b);

            if(a+b==0) break;

            f[a]=b;

            add(b,a);

        }

        int root=1;

        while(f[root])

        {

            root=f[root];

        }

        vis[root]=1;

        dfs(root);

        out(max(dp[root][0],dp[root][1]));

        putchar('\n');

    }

    return 0;

}