text input cond mil accept clas 情況 accepted tar

題意：

一棵有權樹，從根結點中放入 K 個機器人。求用這 K 個機器人遍歷全部的結點最少的權值和。

思路：

1. dp[u][i] 表示給以 u 為根節點的子樹放 i 個機器人，遍歷其子樹所須要的最小權值。

2. 關鍵在於 dp[u][0] 的理解，表示：最後停留在以 u 為根節點的子樹下 0 個機器人，而且遍歷了 u 子樹的最小權值和。

3. 以下的步驟就變成和分組背包類似的情況了，根節點 u 給孩子 v 放多少個機器人。

4. dp[u][i] = min(dp[u][i], dp[u][j] + dp[v][i-j] + (i-j) * c); 能夠理解成給 v 放了 i-j 個機器人，給 v 的其它兄弟放了 j 個，u 總共同擁有 i 個

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

HintIn 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;
}

HDOJ 4003 Find Metal Mineral