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 Specification:
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 Specification:
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
- 题目里虽然说了“It is vitally important to have all the cities connected by highways in a war.”，但他给你的图不一定是一张连通图，甚至还会有孤立顶点，你得把这些图全都连起来
- 用动态申请内存的邻接表会超时，邻接矩阵或静态邻接表不会超时
- 三种思路（假设均使用静态邻接表）：去掉一个顶点并且对剩下的图进行dfs（T=O(k(n+m))，很慢）；去掉一个顶点然后算并查集（T=O(k*m)）；对各子图本别计算其关节点（T=O(n+m)）并在计算过程中顺便算出其划分出的子图数量。不过本题k比较小，前两个方法能过。