Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3532 Accepted Submission(s): 1401

Problem Description

Machines have once again attacked the kingdom of Xions. The kingdom of Xions has N cities and N-1 bidirectional roads. The road network is such that there is a
unique path between any pair of cities.

Morpheus has the news that K Machines are planning to destroy the whole kingdom. These Machines are initially living in K different cities of the kingdom and
anytime from now they can plan and launch an attack. So he has asked Neo to destroy some of the roads to disrupt the connection among Machines. i.e after destroying those roads there should not be any path between any two Machines.

Since the attack can be at any time from now, Neo has to do this task as fast as possible. Each road in the kingdom takes certain time to get destroyed and they
can be destroyed only one at a time.

You need to write a program that tells Neo the minimum amount of time he will require to disrupt the connection among machines.

Input

The first line is an integer T represents there are T test cases. (0<T <=10)
For each test case the first line input contains two, space-separated integers, N and K. Cities are numbered 0 to N-1. Then follow N-1 lines, each containing three, space-separated integers, x y z, which means there is a bidirectional road connecting city x and city y, and to destroy this road it takes z units of time.Then follow K lines each containing an integer. The ith integer is the id of city in which ith Machine is currently located.
2 <= N <= 100,000
2 <= K <= N
1 <= time to destroy a road <= 1000,000

Output

For each test case print the minimum time required to disrupt the connection among Machines.

Sample Input

    15 32 1 81 0 52 4 51 3 4240 
  

Sample Output

    10 
    
     Hint 
    
Neo can destroy the road connecting city 2 and city 4 of weight 5 , and the road connecting city 0 and city 1 of weight 5. As only one road can be destroyed at atime, the total minimum time taken is 10 units of time. After destroying these roads none of the Machines can reach other Machine via any path.

Author

TJU

Source

2012 Multi-University Training Contest 2

问题链接：HDU4313 Matrix

问题描述：（略）

问题分析：

这是一个最小生成树的为问题，解决的算法有Kruskal（克鲁斯卡尔）算法和Prim（普里姆）算法。

程序说明：

本程序使用Kruskal算法实现。有关最小生成树的问题,使用克鲁斯卡尔算法更具有优势，只需要对所有的边进行排序后处理一遍即可。程序中使用了并查集，用来判定加入一条边后会不会产生循环。程序中，图采用边列表的方式存储，排序一下就好了。

这个题给的坑是节点号有可能是0！！！

AC的C++语言程序如下：

/* HDU4313 Matrix */

#include <iostream>
#include <algorithm>
#include <set>
#include <stdio.h>

using namespace std;

const int N = 100000;

int f[N + 1];

void UFInit(int n)
{
    for(int i = 0; i <=n; i++)
        f[i] = i;
}

int Find(int a) {
    return a == f[a] ? a : f[a] = Find(f[a]);
}

bool Union(int a, int b)
{
    a = Find(a);
    b = Find(b);
    if (a != b) {
        f[a] = b;
        return true;
    } else
        return false;
}

struct Edge {
    int u, v;
    int w;
} edges[N];

bool cmp(Edge& a, Edge& b)
{
    return a.w > b.w;
}

set<int> s;

int main()
{
    int t, n, k;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &k);

        for(int i = 0; i < n - 1; i++)
            scanf("%d%d%d", &edges[i].u, &edges[i].v, &edges[i].w);

        s.clear();
        for(int i = 0; i < k; i++) {
            int a;
            scanf("%d", &a);
            s.insert(a);
        }

        UFInit(n);

        // Kruscal算法
        int cnt = 0;
        long long ans = 0;
        sort(edges, edges + n - 1, cmp);
        for(int i = 0; i < n - 1; i++) {
            int u = Find(edges[i].u);
            int v = Find(edges[i].v);
            if(s.count(u) != 0 && s.count(v) != 0) {
                ans += edges[i].w;
                if(++cnt == k - 1)
                    break;
            } else {
                if(s.count(u) != 0)
                    f[v] = u;
                else
                    f[u] = v;
            }
        }

        printf("%lld\n", ans);
    }

    return 0;
}

HDU4313 Matrix【Kruskal算法+并查集】

Matrix

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