HDOJ 4003 Find Metal Mineral

题意：

一棵有权树，从根结点中放入 K 个机器人。求用这 K 个机器人遍历全部的结点最少的权值和。

思路：

1. dp[u][i] 表示给以 u 为根节点的子树放 i 个机器人，遍历其子树所须要的最小权值。

2. 关键在于 dp[u][0] 的理解，表示：最后停留在以 u 为根节点的子树下 0 个机器人，而且遍历了 u 子树的最小权值和。

3. 以下的步骤就变成和分组背包类似的情况了，根节点 u 给孩子 v 放多少个机器人。

4. dp[u][i] = min(dp[u][i], dp[u][j] + dp[v][i-j] + (i-j) * c); 能够理解成给 v 放了 i-j 个机器人，给 v 的其它兄弟放了 j 个，u 总共同拥有 i 个

Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 2354    Accepted Submission(s): 1071

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;

 

Source

The 36th ACM/ICPC
Asia Regional Dalian Site —— Online Contest

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=11000;

struct Edge
{
	int to,next,w;
}edge[2*maxn];

int Adj[maxn],Size;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void Add_Edge(int u,int v,int weight)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].w=weight;
	Adj[u]=Size++;
}

int n,S,K;
int dp[maxn][20];
bool vis[maxn];

void dfs(int u)
{
	vis[u]=true;
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		int w=edge[i].w;
		if(vis[v]) continue;

		dfs(v);

		for(int j=K;j>=0;j--)
		{
			dp[u][j]+=dp[v][0]+2*w;
			for(int jj=0;jj<=j;jj++)
			{
				dp[u][j]=min(dp[u][j],dp[u][j-jj]+dp[v][jj]+jj*w);
			}
		}
	}

}

int main()
{
	while(scanf("%d%d%d",&n,&S,&K)!=EOF)
	{
		init();
		for(int i=1;i<n;i++)
		{
			int a,b,w;
			scanf("%d%d%d",&a,&b,&w);
			Add_Edge(a,b,w);
			Add_Edge(b,a,w);
		}
		memset(dp,0,sizeof(dp));
		memset(vis,0,sizeof(vis));
		dfs(S);
		printf("%d\n",dp[S][K]);
	}
 	return 0;
}