ural 1018-Binary Apple Tree【树状DP】

ゞ浴缸里的玫瑰 2022-07-24 09:06 143阅读 0赞

## 1018. Binary Apple Tree ##

Time limit: 1.0 second  
Memory limit: 64 MB

Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by integers the root of binary apple tree, points of branching and the ends of twigs. This way we may distinguish different branches by their ending points. We will assume that root of tree always is numbered by 1 and all numbers used for enumerating are numbered in range from 1 to *N*, where *N* is the total number of all enumerated points. For instance in the picture below *N* is equal to 5. Here is an example of an enumerated tree with four branches:

As you may know it's not convenient to pick an apples from a tree when there are too much of branches. That's why some of them should be removed from a tree. But you are interested in removing branches in the way of minimal loss of apples. So your are given amounts of apples on a branches and amount of branches that should be preserved. Your task is to determine how many apples can remain on a tree after removing of excessive branches.

### Input ###

First line of input contains two numbers: *N* and *Q* (2 ≤ *N* ≤ 100; 1 ≤ *Q* ≤ *N* − 1). *N* denotes the number of enumerated points in a tree. *Q* denotes amount of branches that should be preserved. Next*N* − 1 lines contains descriptions of branches. Each description consists of a three integer numbers divided by spaces. The first two of them define branch by it's ending points. The third number defines the number of apples on this branch. You may assume that no branch contains more than 30000 apples.

### Output ###

Output should contain the only number — amount of apples that can be preserved. And don't forget to preserve tree's root ;-)

### Sample ###

<table> 
 <tbody> 
  <tr> 
   <th style="vertical-align:top;">input</th> 
   <th style="vertical-align:top;">output</th> 
  </tr> 
  <tr> 
   <td style="border-color:#1a5cc8;vertical-align:top;"> <pre><code>5 2
1 3 1
1 4 10
2 3 20
3 5 20
</code></pre> </td> 
   <td style="border-color:#1a5cc8;vertical-align:top;"> <pre><code>21
</code></pre> </td> 
  </tr> 
 </tbody> 
</table>

**Problem Source:** Ural State University Internal Contest '99 \#2

解题思路：

题目意思：

有一棵苹果树，苹果树的是一棵二叉树，共N个节点，树节点编号为1~N，编号为1的节点为树根，边可理解为树的分枝，每个分支都长着若干个苹果，现在要要求减去若干个分支，保留M个分支，要求这M个分支的苹果数量最多。

输入：

N M

接下来的N-1行是树的边，和该边的苹果数N and M (1 ≤ M < N;1 < N ≤ 100)

输出：

剩余苹果的最大数量。

input

5 2

1 3 1

1 4 10

2 3 20

3 5 20

output

算法：

删除了某个分支，那么这个分支下的子分支也同时删除。

保留M个分支，也就是删除N-M-1个分支。剩余的最多苹果数=总苹果数-剪掉的苹果数。

注意本题给的边并没有按照树根--树叶的形式来给，也没有按照树的顺序给出边。本来想一个节点对应一个分支长着的苹果数量，cost\[v\]就表示v这个节点的苹果数，可以这样做，但是在输入的时候，不知道这个苹果数量是那个节点的，因为不知道哪个是哪个的子结点。所以用了无向图和苹果数加到边上去。

我的解法中：这题的树状DP的整体思想个pku3345是一样的。

有一些不一样的地方要注意一下：

本程序其实不仅仅针对二叉树，可以是任意的树，删除任意个分支都有算。

#include<iostream>
    #include<cstdio>
    #include<cstring> 
    #include<vector>
    #include<limits>
    using namespace std;
    #define MN 110
    int f[2*MN],p[MN],tmp[MN];
    int N,M;
    bool visit[MN];
    struct NODE
    {
    	int val;
    	int cost;
    };
    vector<NODE>G[MN];
    inline int max(int a,int b)
    {
    	return a>b?a:b;
    }
    inline int min(int a,int b)
    {
    	return a<b?a:b;
    }
    void my_clear()
    {
    	int i;
    	for(i=0;i<=N;i++)
    	{
    		G[i].clear();
    	}
    	memset(visit,false,sizeof(visit));
    }
    int DP(int v,int from)
    {
    	visit[v]=true;
    	int i,j,k,s,w,last,now;
    	s=G[v].size();
    	if(s==1)  //这边不再是s==0
    	{
    		p[0]=0;
    		return 1;
    	}
    	last=0;
    	f[from]=0;
    	for(i=0;i<s;i++)
    	{
    		w=G[v][i].val;
    		if(visit[w]==true)
    			continue;
    		now=DP(w,from+last+1);
    		p[now]=p[now-1]+G[v][i].cost;  //这边不要漏,把节点w也给删除
    		for(j=0;j<=last+now;j++)
    			tmp[j]=INT_MAX;
    		for(j=0;j<=last;j++)
    		{
    			for(k=0;k<=now;k++)
    			{
    				tmp[j+k]=min(tmp[j+k],f[from+j]+p[k]);
    			}
    		}
    		last+=now;
    		for(j=0;j<=last;j++)
    		{
    			f[from+j]=tmp[j];
    		}
    	}
    	for(i=0;i<=last;i++)
    		p[i]=f[i+from];
    	last++; //加上自身节点
    	return last;
    }
    int main()
    {
    	int i,a,b,sum,c;
    	NODE tmp;
    	while(scanf("%d%d",&N,&M)!=EOF)
    	{
    		sum=0;
    		my_clear();
    		for(i=1;i<N;i++)
    		{
    			scanf("%d%d%d",&a,&b,&c);
    			tmp.cost=c;
    			tmp.val=b;
    			G[a].push_back(tmp);
    			tmp.val=a;
    			G[b].push_back(tmp);
    			sum+=c;
    		}
    		DP(1,0);
    		printf("%d\n",sum-f[N-M-1]);
    	}
    	return 0;
    }