PAT 1013. Battle Over Cities (25)

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 31 21 31 2 3

Sample Output

100

此题其实就是算图的连通块个数，是属于经典的DFS问题。

#include<iostream>#include<cstring>using namespace std;bool vis[1001]={0};bool break_down[1001]={0};int road[1001][1001]={0};int n,m,k;void dfs(int start){  vis[start]=1;    for(int j=1;j<=n;j++)    {      if(road[start][j]!=0&&vis[j]==0&&break_down[j]==0)       {         vis[j]=1;         dfs(j);       }    }}int main(){  cin>>n>>m>>k;  for(int i=0;i<m;i++)  {    int a,b;    cin>>a>>b;    road[a][b]=road[b][a]=1;  }  for(int i=0;i<k;i++)  {    int con;    cin>>con;    break_down[con]=1;    int count=-1;    for(int i=1;i<=n;i++)    {      if(break_down[i]==0)      {        if(vis[i]==0)        {          dfs(i);          count++;        }      }    }    break_down[con]=0;    memset(vis,0,sizeof(vis));    cout<<count<<endl;  }  return 0;}

我的答案中还定义了break_down数组，存储各个城市的信息，如果为1，那么该城市被破坏，不该被搜到。不过后来也发现，其实跟vis数组功能一样....所以也可以把break_down数组去掉，被破坏的城市先在vis数组中，让它的值为1即可。