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 repaired, 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 repaired, 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 repaired if that city is lost.

Sample Input

3 2 3
1 2
1 3
1 2 3

Sample Output

1
0
0

思路：用而为数组存储图的信息，既哪些城市相连，某个城市被破坏，将数组里所有与它相连的值置为false，然后深度优先遍历此时的图，判断分为了几个独立的部分即可。

脑子卡了一下，数组直接赋值是共享同一个内存的，所以要单独写个函数挨个赋值。代码最后一组数据超时，优化中。。

代码：

import java.io.*;
import java.util.ArrayList;
import java.util.HashSet;
public class Main {
  static int cnum,wnum;
static boolean[][]map;
static boolean[]vis;
static ArrayListcheck = new ArrayList();
static void dfs(boolean[][]m,boolean[]v,int c,ArrayLista){
  for(int i=1;i<=cnum;i++){
    if(!v[i] && m[c][i]){
      v[i] = true;
      a.add(i);
      dfs(m,v,i,a);
    }
  }
}
static int map_num(boolean[][]m,boolean[]v){
  int num = 0;
  ArrayListal = new ArrayList();
  for(int i=1;i<=cnum;i++){
    if(!v[i]){
      al.add(i);
      v[i] = true;
      dfs(m,v,i,al);
      num++;
      if(al.size()>=cnum){
        break;
      }
    }
  }
  return num-2;
}
static void destroy(boolean[][]m,int c){
  for(int i=1;i<=cnum;i++){
    if(i == c)continue;
    m[c][i] = false;
    m[i][c] = false;
  }
}
static boolean[][]copy(boolean[][]m){
  boolean[][]c = new boolean[m.length][m[0].length];
  for(int i=0;i