J - Anniversary Party(30) URAL - 1039

J - Anniversary Party(30) URAL - 1039

Background
The president of the Ural State University is going to make an 80'th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from 1 to N, The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.
Problem
Your task is to make up a guest list with the maximal conviviality rating of the guests.
           

Input

The first line of the input contains a number N. 1 ≤ N ≤ 6000. 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 the supervisor relation tree goes. Each line of the tree specification has the form

<L> <K>

which means that the K-th employee is an immediate supervisor of L-th employee. Input is ended with the line

0 0
           

Output

The output should contain the maximal total rating of the guests.

Example

input output

7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0
           

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
using namespace std;
const int maxn=6005;
vector<int> mq[maxn];
int dp[maxn][2];
void dfs(int cur,int p)
{
    int i,nex;
    for(i=0; i<mq[cur].size(); i++)
    {
        nex=mq[cur][i];
        if(nex==p)
        {
            continue;
        }
        dfs(nex,cur);  //先找從cur這個點可以擴充的dp
        dp[cur][0]+=max(dp[nex][0],dp[nex][1]);
        dp[cur][1]+=dp[nex][0];
    }
}
int main()
{
    int n,i;
    int a,b;
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
        {
            mq[i].clear();
            scanf("%d",&dp[i][1]);
        }
        while(scanf("%d%d",&a,&b)&&(a+b))
        {
            mq[a].push_back(b);
            mq[b].push_back(a);
        }
        dfs(1,0);
        int ans=max(dp[1][0],dp[1][1]);
        printf("%d\n",ans);
    }
    return 0;
}
           

#include<iostream>
#include<cstdio>
#include<map>
#include<math.h>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;

int pre[6010];
vector<int> G[6010]; // 動态存儲鄰接矩陣
int dp[6010][2], vis[6010], n;

void dfs(int u, int father)
{
    int d = G[u].size();
    for(int i = 0; i < d; i++)
    {
        int v = G[u][i];
        if(v != father)
        {
            dfs(v, pre[v] = u);
        }
    }
}

void d(int t)
{
    vis[t] = 1;// 标記已被通路
    for(int i = 1; i <= n; i++)
    {
        if(pre[i] == t && !vis[i])
        {
            d(i);//當周遊到葉子，求最優解
            dp[t][0] += max(dp[i][0], dp[i][1]);
            dp[t][1] += dp[i][0];
        }
    }
}

int main()
{
    int i, u, v;
    while(scanf("%d",&n) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&dp[i][1]);
        }
        while(scanf("%d%d",&u,&v) && (u + v))
        {
            G[u].push_back(v);
            G[v].push_back(u);
        }
        memset(pre, -1, sizeof(pre));//設定初始都指向-1位父親結點
        dfs(1,-1);
        memset(vis, 0, sizeof(vis));
        d(1);
        printf("%d\n",max(dp[1][0], dp[1][1]));
    }
    return 0;
    }
           

#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;
#define N 6005
int dp[N][2], val[N], visit[N];
int in[N];
struct node
{
    int v;
    int next;
} pe[N<<1];
int e, head[N];
void init ()
{
    memset  (head, -1, sizeof (head));
    e = 0;
}
void addedge (int u, int v)
{
    pe[e].v = v;
    pe[e].next = head[u];
    head[u] = e ++;
}
int dfs (int u)
{
    visit[u] = 1;
    dp[u][0] = 0;
    dp[u][1] = val[u];
    for (int i = head[u]; i != -1; i = pe[i].next)
    {
        int v = pe[i].v;
        dfs(v);
        dp[u][0] += max(dp[v][0],  dp[v][1]);
        dp[u][1] += dp[v][0];
    }
    return max(dp[u][0], dp[u][1]);
}
int main()
{
    int n, v, u;
    while (scanf ("%d", &n) != EOF)
    {
        init();
        for (int i = 1; i <= n; ++i)
        {
            in[i] = 0;
            visit[i] = 0;
            scanf ("%d", &val[i]);
        }
        while (scanf ("%d%d", &v, &u) && v + u != 0)
        {
            addedge (u, v);
            in[v]++;
        }
        int ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            if(!in[i])
            {
                ans += dfs(i);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}