Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1390 Accepted Submission(s): 507
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.
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
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
1 5 3 2 1 8 1 0 5 2 4 5 1 3 4 2 4 0
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 a time, 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.
1 //题意：有多个城市，多条路，路都是双向的。有一些城市有机器人大军，我们想通过城市之间的道路破坏机器人大军之间的联系，并且要用最少的时间。 2 //用kruskal算法利用所有的边，生成多棵最大生成树，每棵树上只有一个城市有机器人大军，剩下的边即是需要被破坏的边。 3 #include <iostream> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 8 typedef struct Node 9 { 10 int x, y, v; 11 bool operator <(const Node &a) const 12 { 13 return v > a.v; 14 } 15 }; 16 17 const int N = 100002; 18 Node edge[N]; 19 bool b[N]; 20 int f[N]; 21 22 int find(int x) 23 { 24 if(f[x] == -1) 25 return x; 26 return f[x] = find(f[x]); 27 } 28 29 void fold(int x, int y) 30 { 31 f[find(x)] = find(y); 32 } 33 34 long long kruskal(int n) 35 { 36 int i,j,k; 37 sort(edge, edge +n); //按权由大到小排序 38 long long sum = 0; 39 memset(f, -1, sizeof(f)); 40 for(i = 0; i < n; i ++) 41 { 42 int x = find(edge[i].x); 43 int y = find(edge[i].y); 44 if(b[x] && b[y]) 45 { 46 sum += edge[i].v; 47 continue; 48 } 49 if(b[x] || b[y]) 50 b[x] = b[y] = 1; 51 fold(x,y); 52 } 53 return sum ; 54 } 55 56 int main() 57 { 58 int T, i,j,k; 59 int n, m, x; 60 cin>>T; 61 while(T--) 62 { 63 cin>>n>>m; 64 for(i = 0; i < n-1; i ++) 65 cin>>edge[i].x>>edge[i].y>>edge[i].v; 66 memset(b, 0, sizeof(b)); 67 for(i = 0; i < m; i ++) 68 { 69 cin>>x; 70 b[x] = 1; 71 } 72 cout<<kruskal(n-1)<<endl; 73 } 74 return 0; 75 }
