hdu4003

Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3474    Accepted Submission(s): 1615

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 11 2 11 3 13 1 21 2 11 3 1

 

Sample Output

32
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;
2016年亚洲赛大连站热身赛，当时没有做出来，尴尬，
题意是给一棵树，给一个起点S ，k个机器人，每个机器人只能从起点S出发，最终要使所有点被遍历到，每两个节点之间有一个花费，求出最小的总花费
显然是一个树形dp，
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 <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;const int maxn =1e6+10;const int inf=0x3f3f3f3f;struct node{    int next,v,w;}edge[maxn];int n,s,k;int dp[100000][15];int tot,head[maxn];void init(){    tot=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w){    edge[tot].v=v;    edge[tot].w=w;    edge[tot].next=head[u];    head[u]=tot++;}void dfs(int u,int pre){    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==pre)            continue;            dfs(v,u);        for(int j=k;j>=0;j--)///分组背包        {            dp[u][j]+=dp[v][0]+2*edge[i].w;            for(int l=1;l<=j;l++)            {                dp[u][j]=min(dp[u][j],dp[u][j-l]+dp[v][l]+l*edge[i].w);            }        }    }}int main(){    int u,v,w;    while(scanf("%d%d%d",&n,&s,&k)!=-1)    {        init();        for(int i=1;i<n;i++)        {            scanf("%d%d%d",&u,&v,&w);            addedge(u,v,w);            addedge(v,u,w);        }        memset(dp,0,sizeof(dp));        dfs(s,0);        printf("%d\n",dp[s][k]);    }//cout << "Hello world!" << endl;    return 0;}