Pat grade a a1013 battle over cities

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

The question ：

3 vertices ,2 Two sides , Delete 3 Nodes （ Delete one at a time , All operations are in the original drawing , Deletion does not affect each other ）

 1,2 Have edge ,1,3 Have edge , 1 2 3 Is the deleted node

Delete a node in the graph , Find out how many lines are needed for reconnection in the graph , In fact, it is to calculate a certain vertex , The number of connected blocks , Because after deleting a vertex , Adding a line to two connected blocks can make the graph reconnect .

Ideas ：

There is no need to delete vertices , Just skip it when you encounter it , Then find the number of connected blocks in the graph , The line to connect = = Number of connected blocks - 1;

Code ：

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
int  n, m, k, deleteNode;
const int maxn = 1010;
vector<int > Adj[maxn];
bool vis[maxn] = {false}; 

void DFS(int v, int deleteNode) {
	if(v  == deleteNode) return ;
	for(int i = 0; i < Adj[v].size(); i++) {
		int next = Adj[v][i];
		if(vis[next] == false) {
			vis[next] = true;
			DFS(next,deleteNode);
		}
	}
}

int main () {
	scanf("%d %d %d", &n, &m ,&k);// Number of nodes , The number of sides and the number of nodes to delete ;
	for(int i = 0; i < m; i++) {// Drawing ： 
		int v1, v2;
		scanf("%d %d", &v1, &v2);
		Adj[v1].push_back(v2);
		Adj[v2].push_back(v1);		 
	}
	for(int query = 0; query < k; query++) {
		scanf("%d", &deleteNode);
		memset(vis, false, sizeof(vis));
		int num = 0;
		for(int i = 1; i <= n; i++) {
			if(i != deleteNode && vis[i] == false) {
				DFS(i, deleteNode);
				num++;
			}
		}
		printf("%d\n", num - 1);
	}
	return 0;
}