hdu 3452 Bonsai 求最少去掉多少权值的边使得所有的叶子节点和根节点断开连接

帮助John Kreese通过移除树中特定边的方式确保所有叶子节点与根节点断开连接,达到最小总权重的目标。

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 ui vi wi (where ui, vi ∈ {1,……, n} and 0 <= wi <= 1000) indicating that vertex ui is connected to vertex vi by an undirected edge with weight wi. 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

//


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=3000;
struct Node
{
    int t,w;
    int next;
};
int V;
int root;
Node G[maxn];
int p[maxn];
int l;
void init()
{
    memset(p,-1,sizeof(p));
    l=0;
}
void addedge(int u,int t,int w)
{
    G[l].t=t;
    G[l].w=w;
    G[l].next=p[u];
    p[u]=l++;
}
int du[maxn];
int dfs(int u,int fath)
{
    if(du[u]==1&&u!=root)
    {
        return (1<<29);
    }
    int ans=0;
    for(int i=p[u];i!=-1;i=G[i].next)
    {
        int t=G[i].t,w=G[i].w;
        if(t==fath) continue;
        ans+=min(w,dfs(t,u));
    }
    return ans;
}
int main()
{
    while(scanf("%d%d",&V,&root)==2&&V)
    {
        init();
        memset(du,0,sizeof(du));
        for(int i=0;i<V-1;i++)
        {
            int u,v,w;scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
            du[u]++,du[v]++;
        }
        int ans=dfs(root,-1);
        printf("%d\n",ans);
    }
    return 0;
}

好的,关于 HDU4992 所有原根的问题,我可以给出以下解答: 1. 首先,我们需要知道什么是原根。原根是指模 n 意义下,所有与 n 互质的数 a 的小正整数 k 次幂对 n 取模的值覆盖了所有与 n 互质的数。简单来说,就是如果一个数 a 是模 n 意义下的原根,那么 a 的任何正整数次幂对 n 取模的值都不相同,且能覆盖所有与 n 互质的数。 2. 为了模 n 意义下的所有原根,我们需要先出与 n 互质的数的个数 phi(n)。phi(n) 可以使用欧拉函数出。 3. 然后,我们需要枚举模 n 意义下的所有数,判断它是否是原根。具体来说,对于每个 a,我们需要判断 a 的每个小于 phi(n) 的正整数次幂对 n 取模的值是否都不相同,且能覆盖所有与 n 互质的数。如果是,那么 a 就是模 n 意义下的原根。 4. 代码实现可以参考以下 Java 代码: ``` import java.util.*; public class Main { static int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } static int phi(int n) { int res = n; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { res = res / i * (i - 1); while (n % i == 0) { n /= i; } } } if (n > 1) { res = res / n * (n - 1); } return res; } static int pow(int a, int b, int mod) { int res = 1; while (b > 0) { if ((b & 1) != 0) { res = res * a % mod; } a = a * a % mod; b >>= 1; } return res; } static boolean check(int a, int n, int phi) { for (int i = 1, j = pow(a, i, n); i <= phi; i++, j = j * a % n) { if (j == 1) { return false; } } return true; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); while (scanner.hasNext()) { int n = scanner.nextInt(); int phi = phi(n); List<Integer> ans = new ArrayList<>(); for (int i = 1; i < n; i++) { if (gcd(i, n) == 1 && check(i, n, phi)) { ans.add(i); } } Collections.sort(ans); for (int x : ans) { System.out.print(x + " "); } System.out.println(); } } } ``` 其中,gcd 函数用于大公约数,phi 函数用于欧拉函数,pow 函数用于快速幂模,check 函数用于判断一个数是否是原根。在主函数中,我们依次读入每个 n,出 phi(n),然后枚举模 n 意义下的所有数,判断它是否是原根,将所有原根存入一个 List 中,后排序输出即可。 希望我的回答能够帮到你,如果你有任何问题,欢迎随时提出。
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值