（基础） 树形dp HDU 1520 Anniversary party

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3509    Accepted Submission(s): 1622

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

711111111 32 36 47 44 53 50 0

 

Sample Output

5

 

Source

Ural State University Internal Contest October'2000 Students Session

 

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linle

题意：给出了一个树形结构，每个节点赋予一定的值，可以是负数，一条边上的两个点不能同时取，求一种取法使得sum最大

思路：树形dp  dp[i][0] 表示不取i节点的最大值，dp[i][1]表示取了i节点的最大值  dp[i][0] = sum(max(dp[son][0],dp[son][1]))  

dp[i][1] = sum(dp[son][0])+val[i];

代码：

#include<iostream>

#include<cstdio>

#include<string.h>

#include<algorithm>

#include<string>

#include<deque>

#include<queue>

#include<math.h>

#include<vector>

#include<map>

#include<stack>

#include<set>

using namespace std;

#define MAX 6000+10

#define MOD 99997

const int inf = 0xfffffff;

int dp[MAX][2];

int val[MAX];

vector<int> G[MAX];

bool vis[MAX];

void dfs(int fa,int x)

{

if (vis[x]) return;

vis[x] = true;

for (int i = 0 ; i < G[x].size() ; ++i)

{

int y = G[x][i];

if (y==fa) continue;

dfs(x,y);

dp[x][0] += max(dp[y][0],dp[y][1]);

dp[x][1] += dp[y][0];

}

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

}

int main()

{

#ifdef _DEBUG

freopen("in.txt","r",stdin);

#endif

int n;

while (scanf("%d",&n)==1)

{

memset(vis,false,sizeof(vis));

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

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

{

G[i].clear();

scanf("%d",val+i);

}

int u,v;

while (scanf("%d%d",&u,&v), u || v)

{

G[u].push_back(v);

G[v].push_back(u);

}

int ans = 0;

for (int i = 1 ; i <= n ; ++i) if (!vis[i])

{

dfs(-1,i);

ans += max(dp[i][0],dp[i][1]);

}

printf("%d\n",ans);

}

}