Find Metal Mineral
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 2371 Accepted Submission(s): 1079
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;
dp[pos][num]表示以pos为根节点的子树下，用去num个机器人，所得到的最小值
特别的是当num==0的时候，dp[pos][0]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
状态转移：dp[pos][num]=min∑{dp[pos_j][num_j]+w_j},pos_j是pos的所有儿子，
当num_j==0的时候，和别的时候不同，特别处理一下就好。
状态转移并不难，最精华的，我不认为是状态转移，而是转移时使用的那个“分组背包”思想。
使用一维数组的“分组背包”伪代码如下：
for 所有的组i
for v=V..0
for 所有的k属于组i
f[v]=max{f[v],f[v-c[k]]+w[k]}
上面那个状态转移时，要把num个机器人分给所有它的儿子，状态太多了，不好做，用“分组背包”实在太帅了。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4
5 using namespace std;
6
7 struct Edge{
8 int to;
9 int value;
10 int next;
11 }edge[2*10008];
12 int head[10008],dp[10008][12];
13 int N,S,K,total;
14
15 inline int MIN(int a,int b)
16 {
17 if(a<b) return a;
18 return b;
19 }
20 void addEdge(int start,int end,int value)
21 {
22 edge[total].to=end;edge[total].value=value;
23 edge[total].next=head[start];head[start]=total++;
24
25 edge[total].to=start,edge[total].value=value;
26 edge[total].next=head[end],head[end]=total++;
27 }
28
29 void DP(int source,int pre)
30 {
31 for(int i=head[source];i!=-1;i=edge[i].next)
32 {
33 int to=edge[i].to;
34 if(to==pre) continue;
35 DP(to,source);
36 for(int j=K;j>=0;j--)
37 {
38 dp[source][j]+=dp[to][0]+2*edge[i].value;
39 for(int k=1;k<=j;k++)
40 dp[source][j]=MIN(dp[source][j],dp[source][j-k]+dp[to][k]+k*edge[i].value);
41 }
42 }
43 }
44
45 int main()
46 {
47 while(scanf("%d %d %d",&N,&S,&K)!=EOF)
48 {
49 memset(dp,0,sizeof(dp));
50 memset(head,-1,sizeof(head));
51 total=0;
52
53 int x,y,cost;
54 for(int i=1;i<N;i++)
55 {
56 scanf("%d %d %d",&x,&y,&cost);
57 addEdge(x,y,cost);
58 }
59 DP(S,-1);
60 printf("%d\n",dp[S][K]);
61 }
62 return 0;
63 }
思路和其他人差不多，谈谈自己是怎么理解的吧：
首先，用两数组建立无向树，还是第一次，费了不少功夫。定义一个结构代表边edge[i]={to,next,value},其中edge[i]表示第i条边，to是边的末点，value就不说了，next代表的是同一父节点边方向相同的与此边相邻的左边的边的编号，其中最左是-1。以此为基础建立一个边数组（注意数组大小是顶点数的两倍），还有用一个head[i]数组来表示最右边的边的编号。
解释这一段代码，当时写的理解的不是很好：
1 void DP(int source,int pre)
2 {
3 for(int i=head[source];i!=-1;i=edge[i].next)
4 {
5 int to=edge[i].to;
6 if(to==pre) continue;
7 DP(to,source);
8 for(int j=K;j>=0;j--)
9 {
10 dp[source][j]+=dp[to][0]+2*edge[i].value;
11 for(int k=1;k<=j;k++)
12 dp[source][j]=MIN(dp[source][j],dp[source][j-k]+dp[to][k]+k*edge[i].value);
13 }
14 }
15 }
第一层for循环查找与节点source相连的各条边，to是此边对应的末点，若此边是叶子节点则to==pre（由于节点最左边的边访问完以后，只有到他父节点的边，这里只能用continue，因为不确定访问的顺序，return不合适）,递归访问子节点此时应用01背包，由于dp[source][0]比较特殊单独处理，一棵子树对应一个分组，dp[source][j]=MIN(dp[source][j],dp[source][j-k]+dp[to][k]+k*edge[i].value);也比较好理解。
树形DP-----HDU4003 Find Metal Mineral