poj 2342 Anniversary party 簡單樹形dp

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3862		Accepted: 2171

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

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

【題意】

公司有n個人，每個人有價值vi，有一天舉辦年會，每個人都可以參加，但有嚴格的等級制度，參加活動時，不能同時出現a和a的上司，問如何才能使總和最大。

【分析】

每個人只有去和不去兩種狀態，設DP[i][0]和DP[i][1]分別表示第i個人不參加和參加年會，獲得的總的最大價值。

則狀態轉移方程爲：

DP[i][1] += DP[j][0],

DP[i][0] += max{DP[j][0],DP[j][1]};其中j爲i的孩子節點。

這樣，從根節點r進行dfs，最後結果爲max{DP[r][0],DP[r][1]}。

（分析來自yzmduncan）

第2~n+1行爲這n個人的價值

#include "stdio.h"   //簡單的樹形dp題
#include "string.h"
#include "queue"
using namespace std;

#define N 60005
#define INF 0x3fffffff

struct node
{
    int x,y;
    int weight;
    int next;
}edge[4*N];
int idx,head[N];

int root;
int du[N];
int value[N];

int dp[N][2];
int MAX(int a,int b) { return a>b?a:b; }

void Init()
{
    idx = 0;
    memset(head,-1,sizeof(head));
}

void Add(int x,int y,int weight)
{
    edge[idx].x = x;
    edge[idx].y = y;
    edge[idx].weight = weight;
    edge[idx].next = head[x];
    head[x] = idx++;
}

void DFS(int i)  //
{
    int k,j;
    dp[i][0] = 0;
    dp[i][1] = value[i];
    for(k=head[i]; k!=-1; k=edge[k].next)
    {
        j = edge[k].y;
        DFS(j);
        dp[i][0] += MAX(dp[j][0],dp[j][1]);
        dp[i][1] += dp[j][0];
    }
}


int main()
{
    int n;
    int i;
    int x,y;
    while(scanf("%d",&n)!=EOF)
    {
        Init();
        memset(du,0,sizeof(du));  //記錄節點的入度
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; ++i)
            scanf("%d",&value[i]);
        while(scanf("%d %d",&x,&y) && x+y>0)
        {
            Add(y,x,0);
            du[x]++;
        }
        for(i=1; i<=n; ++i)
        {
            if(du[i]==0)
                root = i;
        }
        DFS(root);
        printf("%d\n",MAX(dp[root][0],dp[root][1]));
    }
    return 0;
}