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POJ 2342 簡單樹形dp

編輯：C++入門知識

Anniversary party Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3719 Accepted: 2080

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 N – 1 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

Sample Output

子節點與父節點不能同時選，求權值，簡單樹形dp

dp[u][0]表示當前節點不選，dp[u][1]表示當前節點選。

狀態方程：dp[u][0]+=max(dp[v][0],dp[v][1])表示當前節點不選時疊加子節點選與不選的最大值。

dp[u][1]+=dp[v][0]，表示當前節點選時，子節點不選。

代碼：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-6 3:11:18
File Name :3.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=10001;
int head[maxn],weight[maxn],dp[maxn][2],tol;
struct node{
	int next,to;
}edge[2*maxn];
void add(int u,int v){
	edge[tol].to=v;
	edge[tol].next=head[u];
	head[u]=tol++;
}
void dfs(int u,int fa){
	dp[u][1]=weight[u];
    for(int i=head[u];i!=-1;i=edge[i].next){
	    int v=edge[i].to;
		if(v==fa)continue;
		dfs(v,u);
		dp[u][1]+=dp[v][0];
		dp[u][0]+=max(dp[v][1],dp[v][0]);
	}
}
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int i,j,k,m,n;
	 while(~scanf("%d",&n)){
		 memset(head,-1,sizeof(head));tol=0;
		 memset(dp,0,sizeof(dp));
		 for(i=1;i<=n;i++)
			 scanf("%d",&weight[i]);
		 while(~scanf("%d%d",&i,&j)&&i&&j){
			 add(i,j);
			 add(j,i);
		 }
		 dfs(1,-1);
		 //cout<<"hahha"<