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​

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

Code

#include <iostream> 
#include <algorithm> 
#include <cstdio> 
using namespace
 
 std; 
int v[1001][1001];//記錄連通路線 預設為0 不連通 
bool visit[1001];//記錄是否遍歷過 
int n;//n個城市 
void dfs(int node)
{ 
	visit[node] = true; 
	for(int i=1; i<=n; i++)
	{ 
		if(visit[i]==false && v[node][i] == 1)
		{ 
			dfs(i); 
		} 
	} 
} 
int main() 
{ 
	int m, k, a, b;//m條路 k個要檢查的城市 ab為路線起點終點 
	scanf("%d%d%d", &n, &m, &k); 
	for(int i=0; i<m; i++)
	{ 
		scanf("%d%d", &a, &b); 
		v[a][b] = 1; 
		v[b][a] = 1; 
	} 
	for(int i=0; i<k; i++)
	{ 
		fill(visit, visit+1001, false);//重置visit 所有城市未被遍歷 
		int temp = 0; 
		scanf("%d", &temp); 
		visit[temp] = true;//被攻佔的城市，標記為true 
		int cnt = 0;//記錄連通分量 
		for(int j=1; j<=n; j++)
		{ 
			if(visit[j] == false)
			{ 
				dfs(j); 
				cnt++;//連通分量+1 
			} 
		}
		printf("%d\n", cnt-1);//思路第二條 
	} 
	return 0; 
}

思路

以上程式碼來自 https://blog.csdn.net/cv_jason/article/details/81019874 本來照著他的程式碼自己打了一遍，但是在最後超時了，原因是我用了cin和cout，而他用的是scanf和printf，發現這個原因不是很震驚，因為之前也聽ACM大佬說cin和cout比scanf和printf慢很多，但是，因為懶，所以一直用cin和cout。。。

以上