Battle Over Cities (25)（DFS、连通图）

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

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

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

主要思想：一张连通图，去掉一个节点后，想把个个节点连通起来，至少需要添加的边的数目n。n = 该连通图，去掉一个节点后，所形成的 极大连通子图的个数 - 1.

所以 只要把去掉的点定义为检测点，DFS不要去遍历它，计算出极大连通子图的个数便可，算法如下：

 1    void DFS(int x,int N) //N为检测点
 2
 3 {
 4
 5       visit[x]=1;
 6
 7     for(int i=0;i<Ladj[x].size();i++)
 8
 9       {
10
11             if(visit[Ladj[x][i]]==0&&Ladj[x][i]!=N)//不等于被检测的点
12
13                   DFS(Ladj[x][i],N);
14
15       }
16
17 }

 1 int  count =0；
 2
 3       for(j=1;j<=n;j++)
 4
 5             {
 6
 7       if(visit[j]==0&&j!=check[i])//不等于被检测的点，check数组保存检测点
 8
 9               {
10
11                  count++;//极大连通子图的个数
12
13                   DFS(j,check[i]);
14
15               }
16
17             }
18
19
20
21             cout<<count-1<<endl;
22
23  

AC代码：

  1 #include <iostream>
  2
  3 #include <vector>
  4
  5 using namespace std;
  6
  7
  8
  9
 10
 11 vector<int> Ladj[1001];
 12
 13
 14
 15 int check[1001];//存放要检测的点
 16
 17
 18
 19 int visit[1001];
 20
 21
 22
 23 void DFS(int x,int N)
 24
 25 {
 26
 27       visit[x]=1;
 28
 29     for(int i=0;i<Ladj[x].size();i++)
 30
 31       {
 32
 33             if(visit[Ladj[x][i]]==0&&Ladj[x][i]!=N)//不等于被检测的点
 34
 35                   DFS(Ladj[x][i],N);
 36
 37       }
 38
 39 }
 40
 41
 42
 43 int main()
 44
 45 {
 46
 47
 48
 49       int n,m,k,i,j;
 50
 51       while(cin>>n)
 52
 53       {
 54
 55         cin>>m>>k;
 56
 57
 58
 59
 60
 61
 62
 63         for(i=0;i<m;i++)
 64
 65         {
 66
 67            int a,b;
 68
 69             cin>>a>>b;
 70
 71          Ladj[a].push_back(b);
 72
 73             Ladj[b].push_back(a);
 74
 75         }
 76
 77
 78
 79          for(i=0;i<k;i++)
 80
 81         {
 82
 83           cin>>check[i];
 84
 85         }
 86
 87
 88
 89           for(i=0;i<k;i++)
 90
 91         {
 92
 93             int count=0;
 94
 95           for(j=1;j<=n;j++)//初始化
 96
 97             {
 98
 99               visit[j]=0;
100
101             }
102
103
104
105             for(j=1;j<=n;j++)
106
107             {
108
109               if(visit[j]==0&&j!=check[i])//不等于被检测的点
110
111               {
112
113                  count++;
114
115                   DFS(j,check[i]);
116
117               }
118
119             }
120
121
122
123             cout<<count-1<<endl;
124
125
126
127         }
128
129       }
130
131       return 0;
132
133 }