PAT 甲级 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

题意：

3个顶点，2个边长，删除3个结点（每次删一个，都是在原图内操作，删除互不影响）

 1，2 有边，1，3有边， 1 2 3 是删除的结点

在图中删除某个结点，求图内重新连通需要多少根线，其实就是求不计某个顶点，连通块的个数，因为在删除某顶点后，两两连通块内加一根线就能使得图重新连通了。

思路：

没必要真的把顶点删除，遇到就跳过就好了，这样再去找图内连通块的个数，连通起来要用的线 = = 连通块个数 - 1；

代码：

#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);//结点个数，边数和要删除的结点个数;
	for(int i = 0; i < m; i++) {//建图： 
		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;
}