HDU4003:Find Metal Mineral(树形DP)

Problem Description

Humans have discovered a kind of new metal mineral on Mars which are distributed in point‐like with paths connecting each of them which formed a tree. Now Humans launches k robots on Mars to collect them, and due to the unknown reasons, the landing site S of all robots is identified in advanced, in other word, all robot should start their job at point S. Each robot can return to Earth anywhere, and of course they cannot go back to Mars. We have research the information of all paths on Mars, including its two endpoints x, y and energy cost w. To reduce the total energy cost, we should make a optimal plan which cost minimal energy cost.

 

Input

There are multiple cases in the input.  
In each case:  
The first line specifies three integers N, S, K specifying the numbers of metal mineral, landing site and the number of robots.  
The next n‐1 lines will give three integers x, y, w in each line specifying there is a path connected point x and y which should cost w.  
1<=N<=10000, 1<=S<=N, 1<=k<=10, 1<=x, y<=N, 1<=w<=10000.

 

Output

For each cases output one line with the minimal energy cost.

 

Sample Input

   
   
   
   

    
    
    
    3 1 1
1 2 1
1 3 1
3 1 2
1 2 1
1 3 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3
2


    
    
    
    
 
     
 
      Hint 
     
In the first case: 1->2->1->3 the cost is 3; In the second case: 1->2; 1->3 the cost is 2; 
    
    
    
    
     
   
   
   
   

 

 

题意：给出结点数n,起点s，机器人数k，然后n-1行给出相互连接的两个点，还有这条路线的价值，要求最小花费

思路：这是我从别人博客里找到的解释，因为很详细就引用了

dp[i][j]表示对于以i结点为根结点的子树，放j个机器人所需要的权值和。 

当j=0时表示放了一个机器人下去，遍历完结点后又回到i结点了。状态转移方程类似背包 

如果最终的状态中以i为根结点的树中有j(j>0)个机器人，那么不可能有别的机器人r到了这棵树后又跑到别的树中去 

因为那样的话，一定会比j中的某一个到达i后跑与r相同的路径再回到i，再接着跑它的路径要差(多了一条i回去的边) 

这样的话，如果最后以i为根结点的树中没有机器人，那么只可能是派一个机器人下去遍历完后再回来

 

可以这么理解：

对于每个根节点root，有个容量为K的背包

如果它有i个儿子，那么就有i组物品，价值分别为dp[son][0],dp[son][1].....dp[son][k] ，这些物品的重量分别为0,1,.....k

现在要求从每组里选一个物品（且必须选一个物品）装进root的背包，使得容量不超过k的情况下价值最大。

那么这就是个分组背包的问题了。

但是这里有一个问题，就是每组必须选一个物品。

对于这个的处理，我们先将dp[son][0]放进背包，如果该组里有更好的选择，那么就会换掉这个物品，否则的话这个物品就是最好的选择。这样保证每组必定选了一个。

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct Node
{
    int now,next,val;
}tree[20005];

int dp[10005][15];
int head[10005];
int n,s,k,len;

void add(int x,int y,int w)//建树
{
    tree[len].now = y;
    tree[len].val = w;
    tree[len].next = head[x];
    head[x] = len++;
}

void dfs(int root,int p)
{
    int i,j,l,son;
    for(i = head[root];i!=-1;i = tree[i].next)
    {
        son = tree[i].now;
        if(son == p)
        continue;
        dfs(son,root);
        for(j = k;j>=0;j--)
        {
            dp[root][j]+=dp[son][0]+2*tree[i].val;//先将dp[son][0]放进背包，由于dp[son][0]是表示用一个机器人去走完所有子树，最后又回到pos这个节点，所以花费要乘以2
            for(l = 1;l<=j;l++)//在这里找到比dp[son][0]更优的方案，分组背包
            dp[root][j] = min(dp[root][j],dp[root][j-l]+dp[son][l]+l*tree[i].val);
        }
    }
}

int main()
{
    int i,x,y,w;
    while(~scanf("%d%d%d",&n,&s,&k))
    {
        len = 0;
        memset(head,-1,sizeof(head));
        memset(dp,0,sizeof(dp));
        for(i = 1;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,w);
            add(y,x,w);
        }
        dfs(s,-1);
        printf("%d\n",dp[s][k]);
    }

    return 0;
}