HDU 3452 Bonsai（树形dp）

Bonsai

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 33   Accepted Submission(s) : 13

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Problem Description

After being assaulted in the parking lot by Mr. Miyagi following the "All Valley Karate Tournament", John Kreese has come to you for assistance. Help John in his quest for justice by chopping off all the leaves from Mr. Miyagi's bonsai tree!
You are given an undirected tree (i.e., a connected graph with no cycles), where each edge (i.e., branch) has a nonnegative weight (i.e., thickness). One vertex of the tree has been designated the root of the tree.The remaining vertices of the tree each have unique paths to the root; non-root vertices which are not the successors of any other vertex on a path to the root are known as leaves.Determine the minimum weight set of edges that must be removed so that none of the leaves in the original tree are connected by some path to the root.

Input

The input file will contain multiple test cases. Each test case will begin with a line containing a pair of integers n (where 1 <= n <= 1000) and r (where r ∈ ｛1,……, n｝) indicating the number of vertices in the tree and the index of the root vertex, respectively. The next n-1 lines each contain three integers u_i v_i w_i (where u_i, v_i ∈ {1,……, n} and 0 <= w_i <= 1000) indicating that vertex u_i is connected to vertex v_i by an undirected edge with weight w_i. The input file will not contain duplicate edges. The end-of-file is denoted by a single line containing "0 0".

Output

For each input test case, print a single integer indicating the minimum total weight of edges that must be deleted in order to ensure that there exists no path from one of the original leaves to the root.

Sample Input

15 15
1 2 1
2 3 2
2 5 3
5 6 7
4 6 5
6 7 4
5 15 6
15 10 11
10 13 5
13 14 4
12 13 3
9 10 8
8 9 2
9 11 3
0 0

Sample Output

16

Source

2009 Stanford Local ACM Programming Contest

 

题意：

形成一颗以r为根的树，让所有的叶子节点和r节点没有联系，切割一些边，使切割的边的值之和最小。

题解：

dp[i]表示以i节点为根节点，让叶子节点和i节点分开的最小代价。

#include <iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
struct node
{
    int x,w;
    node(int a,int b){x=a; w=b;}
};
int n,r;
vector<node> mp[1005];
int dp[1005];
void dfs(int k,int fa,int w)
{
    dp[k]=w;
    int sum;
    if (mp[k].size()==1 && k!=r) sum=1000000;//如果已经到叶子节点了，那么要赋值最大，后面求出来的值在比较中才能有效。
      else sum=0;
   for(int i=0;i<mp[k].size();i++)
   {
       if (mp[k][i].x==fa) continue;
       dfs(mp[k][i].x,k,mp[k][i].w);
       sum+=dp[mp[k][i].x];
   }
   //printf("%d: %d
",k,sum);
   dp[k]=min(dp[k],sum);
}
int main()
{
    while(~scanf("%d%d",&n,&r))
    {
        if (n==0 && r==0) break;
        for(int i=1;i<=n;i++) mp[i].clear();
        for(int i=1;i<n;i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            mp[x].push_back(node(y,w));
            mp[y].push_back(node(x,w));
        }
        memset(dp,0,sizeof(dp));
        dfs(r,-1,1000000);
        printf("%d
",dp[r]);
    }

    return 0;
}