1013 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 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：被敌军占领的城市的道路全部失效，没有被占领的城市之间需要修复多少条路才能联通。

思路：求有多少个连通图。用dfs。通过某种数据结构来存储他们之间的关系。这里用cin居然被卡了。

#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int maxn = 1e3+10;
int e[maxn][maxn];
bool vis[maxn]; 
int n,m,k;

void dfs(int idx)
{
	for(int i = 1;i <= n;++i)
	{
		if(!vis[i] && e[idx][i])
		{
			 vis[i] = 1;
			 dfs(i);
		}
	}
} 
 
int main(){
	
	
	scanf("%d%d%d",&n,&m,&k);
	memset(e,0,sizeof(e));
	
	for(int i = 0;i < m;++i)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		e[x][y] = e[y][x] = 1; //无向图 
	}
	
 	while(k--)
	{
		memset(vis,0,sizeof(vis));
		int d,cnt = 0;
		scanf("%d",&d);
		vis[d] = 1; //标记
		for(int i = 1;i <= n;++i)
		{
			if(!vis[i])
			{
				vis[i] = 1;
				cnt++;  //几个连通图 
				dfs(i);
			}
		}
		cout << cnt-1 << endl;
	} 
    return 0; 
}

当然我们可以优化这里，可以发现dfs里面的for，可以优化，我们不需要遍历n次，只需要遍历我们关联的边。可以使用vector。

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

const int maxn = 1e3+10;
vector vt[maxn];
bool vis[maxn]; 
int n,m,k;

void dfs(int idx)
{
	vector::iterator t;
	for( t = vt[idx].begin(); t != vt[idx].end();t++)
	{
		if(!vis[*t])
		{
			vis[*t] = 1;
			dfs(*t);
		}
	}
} 
 
int main(){
	
	
	scanf("%d%d%d",&n,&m,&k);

	
	for(int i = 0;i < m;++i)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		vt[x].push_back(y);
		vt[y].push_back(x);
	}
	
 	while(k--)
	{
		memset(vis,0,sizeof(vis));
		int d,cnt = 0;
		scanf("%d",&d);
		vis[d] = 1; //标记
		for(int i = 1;i <= n;++i)
		{
			if(!vis[i])
			{
				vis[i] = 1;
				cnt++;  //几个连通图 
				dfs(i);
			}
		}
		cout << cnt-1 << endl;
	} 
    return 0; 
}