1013 Battle Over Cities （25 分）【连通分量+dfs】

1013 Battle Over Cities （25 分）

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 city​1​​-city​2​​ and city​1​​-city​3​​. Then if city​1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2​​-city​3​​.

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

题意：给你n个城市，m条路，做k次处理，每一次就这个城市被敌人占据了，就要恢复剩余的城市保持连通，问最少需要修几条路。

解题思路： 其实就是那个点删除后剩下的点的连通分量减一，连通分量用dfs来完成，具体的见代码

#include
using namespace std;
int road[1010][1010];
int n;
bool visit[1010];
void dfs(int node)
{
	visit[node]=true;
	for(int i=1;i<=n;i++)
	{
		if(visit[i]==false&&road[node][i]==1)
		dfs(i);		
	}	
} 
int main(void)
{
	int m,k;
	scanf("%d %d %d",&n,&m,&k);
	for(int i=0;i