HDU4313Matrix(用最小生成树思想，集合划分)

Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2598    Accepted Submission(s): 973

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.
 

 解析：要求给出的k个点分开，因为任意两个集合点只有一条相连，所以我们把这k个点当成k个集合点的根节点，那么连接这k个集合的边，它们的权值加起来就是我们要的最小和了。我们将边按从大到小的顺序排好序，然后就是看是否这条边能够使得两个同类的节点连在一起，如果能够的话，那么这条边就是我们要选择的划分边。

#include<stdio.h>#include<algorithm>using namespace std;#define LL __int64const int N = 100005;struct EDG{    int u,v;    LL c;};int fath[N],typeNode[N],n;EDG edg[N];void init(){    for(int i=0;i<=n;i++)    {        fath[i]=i; typeNode[i]=0;    }}int findfath(int x){    if(x!=fath[x])    fath[x]=findfath(fath[x]);    return fath[x];}LL setfath(int i){    int a=findfath(edg[i].u);    int b=findfath(edg[i].v);    if(a==b)    return 0;    if(typeNode[a]&&typeNode[b])    return edg[i].c;    if(typeNode[a])    fath[b]=a;    else    fath[a]=b;    return 0;}int cmp(EDG a,EDG b){    return a.c>b.c;}int main(){    int t,k,a;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        init();        for(int i=0;i<n-1;i++)        scanf("%d%d%I64d",&edg[i].u,&edg[i].v,&edg[i].c);        sort(edg,edg+n-1,cmp);        while(k--)        {            scanf("%d",&a);            typeNode[a]=1;        }        LL ans=0;        for(int i=0;i<n-1;i++)        ans+=setfath(i);        printf("%I64d\n",ans);    }}