1013. Battle Over Cities (25)

1013. Battle Over Cities (25)
It is vitally important to have all the cities connected by highways in a war. if a city is occupied by the enemy, all the highways from/toward that city are closed. we must know immediately if we need to repair any other highways to keep the rest of the cities connected. given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be retried red, quickly.

For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway responred, that is the highway city2-city3.

Input

Each input file contains one test case. each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. the cities are numbered from 1 to N. finally there is a line containing K numbers, which represent the cities we concern.

Output

For each of the K cities, output in a line the number of highways need to be retried red if that city is lost.

Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0

0

# Include
 
  
# Include
  
   
Using namespace std; # include
   
    
# Define N 1001int main () {int n, m, k, I, j, t, c1, c2; bool highways [N] [N]; int concern [N]; memset (highways, 0, sizeof (highways); memset (concern, 0, sizeof (concern); cin> n; cin> m; cin> k; for (I = 0; I
    
     
> C1; cin> c2; highways [c1] [c2] = true; highways [c2] [c1] = true;} for (I = 0; I
     
      
> Concern [I];} for (I = 0; I
      
        Q; memset (visit, 0, sizeof (visit); visit [concern [I] = 1; count = 0; for (j = 1; j <= n; j ++) {if (visit [j]! = 1) {queue
       
         Dq; visit [j] = 1; dq. push (j); // perform BFS while (! Dq. empty () {tmp = dq. front (); dq. pop (); for (t = 0; t <= n; t ++) {if (highways [tmp] [t]! = False & visit [t]! = 1) {dq. push (t); visit [t] = 1 ;}}/ * if an existing node can traverse all vertices, Edge Addition is not required and exit. Otherwise, the independent "island" + 1 searches for the next island */count ++ ;}} cout through the next node that is directly connected and not accessed. <