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poj2342 Anniversary party 簡單樹形dp

編輯：C++入門知識

poj2342 Anniversary party 簡單樹形dp

http://poj.org/problem?id=2342

Anniversary party Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4322 Accepted: 2459

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

Source

Ural State University Internal Contest October'2000 Students Session

題意：每個人都有個價值，然後每個人不和他的直系上屬下屬一起，問這樣產生的最大價值多少。。

題解：入門級的tree dp,dp[i][0]表示以i為根的子樹不選i時產生的最大價值，dp[i][1]表示選i產生的。轉移是:

dp[i][0]=sum(max(dp[j][0],dp[j][1]))

dp[i][1]+=sum(dp[j][0]) (i是j的父親)

邊界是dp[i][0]=0,dp[i][1]=value[i].

結果就是max(dp[root][0],dp[root][1])

加的是雙向邊，這樣任選一個作為最開始的root都可以

代碼:

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=6006;
struct node{
    int to,next;
}e[N*2];
int head[N];
int tot;
int dp[N][2];
bool vis[N];
void init()
{
    tot=0;
    clr1(head);
}
void add(int u,int v)
{
    e[tot].to=v;
    e[tot].next=head[u];
    head[u]=tot++;
}
void dfs(int u)
{
    vis[u]=true;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].to;
        if(!vis[v])
        {
            dfs(v);
            dp[u][0]+=max(dp[v][0],dp[v][1]);
            dp[u][1]+=dp[v][0];
        }
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&dp[i][1]);
            dp[i][0]=0;
            vis[i]=false;
        }
        int u,v;
        for(int i=1;i