1013 Battle Over Cities （DFS求图的连通分量个数）

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

思路：
　　1、给出了城市之间的连通情况，自然想到是邻接矩阵。
　　2、去掉某个城市，问添加几条边能使它们互相连通，也就是求连通分量了。
　　3、我最初使用cin和cout输入输出，超时，改为scanf和printf时，能过，不过时间复杂度还是很高。
　　4、这个是无向图，所以在赋值时必须M[x][y]和M[y][x]都要赋值。

#include<iostream>
#include<sstream>
#include<algorithm>
#include<map>
#include<cmath>
#include<cstdio>
#include<string.h>
using namespace std;
int M[1001][1001];
void dfs(int v0,int visited[],int n)
{
    for(int i=1;i<=n;i++)
    {
        if(visited[i]==0&&M[v0][i]==1)
        {
            visited[i]=1;
            dfs(i,visited,n);
        }
    }
}

int main()
{
    int n,m,k;
    scanf("%d %d %d",&n,&m,&k);
    int visited[n+1];
    memset(visited,0,sizeof(visited));
    memset(M,0,sizeof(M));
    int a[k];

    for(int i=0;i<m;i++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        M[x][y]=1;
        M[y][x]=1;
    }
    for(int i=0;i<k;i++)
    {
        cin>>a[i];
    }
    for(int i=0;i<k;i++)
    {
        int temp=a[i];
        visited[temp]=1;
        int t=0;
        for(int j=1;j<=n;j++ )
        {
            if(visited[j]==0)
            {
                dfs(j,visited,n);
                t++;
            }

        }
        printf("%d
",t-1);
        memset(visited,0,sizeof(visited));

    }
    return 0;
}