hdu 5242 树链剖分找权值最大的前k条链

http://acm.hdu.edu.cn/showproblem.php?pid=5242

Problem Description

It is well known that Keima Katsuragi is The Capturing God because of his exceptional skills and experience in ''capturing'' virtual girls in gal games. He is able to play  k  games simultaneously.

One day he gets a new gal game named ''XX island''. There are  n  scenes in that game, and one scene will be transformed to different scenes by choosing different options while playing the game. All the scenes form a structure like a rooted tree such that the root is exactly the opening scene while leaves are all the ending scenes. Each scene has a value , and we use  wi  as the value of the  i -th scene. Once Katsuragi entering some new scene, he will get the value of that scene. However, even if Katsuragi enters some scenes for more than once, he will get  wi  for only once.

For his outstanding ability in playing gal games, Katsuragi is able to play the game  k  times simultaneously. Now you are asked to calculate the maximum total value he will get by playing that game for  k  times.

 

Input

The first line contains an integer  T ( T≤20 ), denoting the number of test cases.

For each test case, the first line contains two numbers  n,k(1≤k≤n≤100000) , denoting the total number of scenes and the maximum times for Katsuragi to play the game ''XX island''.

The second line contains  n  non-negative numbers, separated by space. The  i -th number denotes the value of the  i -th scene. It is guaranteed that all the values are less than or equal to  231−1 .

In the following  n−1  lines, each line contains two integers  a,b(1≤a,b≤n) , implying we can transform from the  a -th scene to the  b -th scene.

We assume the first scene(i.e., the scene with index one) to be the opening scene(i.e., the root of the tree).

 

Output

For each test case, output ''Case #t:'' to represent the  t -th case, and then output the maximum total value Katsuragi will get.

 

Sample Input

   
   
   
   

    
    
    
    2
5 2
4 3 2 1 1
1 2
1 5
2 3
2 4
5 3
4 3 2 1 1
1 2
1 5
2 3
2 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 10
Case #2: 11
   
   
   
   
   
   
   
   


   
   
   
   

/**
hdu 5242  树链剖分找权值最大的前k条链
题目大意：给定一个树形的游戏网络，可以从根节点出发k个人，每个人可以沿着一条路径走下去，不能回头，出口在各个叶子节点，在路过一个节点时可以
          获得该点的权值，每个点的权值只能被获得一次，问k个人怎样走最后可以获得的权值最多
解题思路：首先从反向建立一棵有向树（从叶子节点到根节点），首先dfs1找出每个节点到根节点的最大权路径，然后按权值递减排序，dfs2找每个点到根节点
          的最大权路径，走过的点不能重复走，最后在这些最大权路径中取前k大即为答案
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100005;
typedef long long LL;
int n,m;
LL a[maxn],x[maxn];
int head[maxn],ip;
bool vis[maxn];

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

struct note
{
    int v,next;
}edge[maxn*2];

struct node
{
    LL sum;
    int id;
    bool operator <(const node &other)const
    {
        return sum>other.sum;
    }
}p[maxn];

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u],head[u]=ip++;
}

LL dfs1(int u)
{
    if(vis[u])return p[u].sum;
    vis[u]=1;
    p[u].sum=a[u];
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        p[u].sum+=dfs1(v);
    }
    return p[u].sum;
}
LL dfs2(int u)
{
    if(vis[u])return 0;
    vis[u]=1;
    LL x=a[u];
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        x+=dfs2(v);
    }
    return x;
}
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
            p[i].id=i;
        }
        init();
        for(int i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            addedge(y,x);
        }
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            if(!vis[i])
                dfs1(i);
        }
        sort(p+1,p+n+1);
        /*for(int i=1;i<=n;i++)
        {
            printf("%d:%d\n",p[i].id,p[i].sum);
        }*/
        memset(x,0,sizeof(x));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            x[i]=dfs2(p[i].id);
        }
        sort(x+1,x+n+1);
        LL ans=0;
        for(int i=n;i>=1&&i>n-m;i--)
        {
            ans+=x[i];
        }
        printf("Case #%d: %I64d\n",++tt,ans);
    }
    return 0;
}