HDU 4003

Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3368    Accepted Submission(s): 1569

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

 

题意：

　　k个机器人从同一地点出发，求走完所有路的最小花费

思路：

　　分组背包，

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。

但是这里有一个问题，就是每组必须选一个物品。

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。

 

dp[i][j] 表示以i为根节点子树，去j个机器人花费的最小值。

 

知识

分组背包：

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]}

AC代码：

 

# include <bits/stdc++.h>
using namespace std;
const int MAX = 10010;
struct node
{
    int to;
    int val;
    int next;
}tree[MAX * 2];
int head[MAX];
int tol;
int dp[MAX][15];
void add(int a, int b, int val)
{
    tree[tol].to = b;
    tree[tol].val = val;
    tree[tol].next = head[a];
    head[a] = tol++;
}
int n, s, k;
void dfs(int root, int f)
{
    for(int i = head[root]; i != -1; i = tree[i].next)
    {
        int son = tree[i].to;
        if(son == f)
            continue;
        dfs(son, root);
        for(int j = k; j >= 0; j--)
        {
            dp[root][j] += dp[son][0] + 2 * tree[i].val; //先将dp[son][0]放进背包
            for(int l = 1; l <= j; l++)
            {
                dp[root][j] = min(dp[root][j], dp[root][j - l] + dp[son][l] + l * tree[i].val);
                // 更优的方案
            }
        }
    }
}
int main()
{
    while(scanf("%d %d %d", &n, &s, &k) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        memset(head, -1, sizeof(head));
        tol = 0;
        for(int i = 1; i < n; i++)
        {
            int a, b, val;
            scanf("%d %d %d", &a, &b, &val);
            add(a, b, val);
            add(b, a, val);
        }
        dfs(s, -1);
        printf("%d
", dp[s][k]);
    }
    return 0;
}