1013. Battle Over Cities (25)

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 3
1 2
1 3
1 2 3
Sample Output
1
0

0

#include
#include
using namespace std;
#include
#define  N 1001
int main()
{
    int n,m,k,i,j,t,c1,c2;
    bool highways[N][N];
    int concern[N];
    memset(highways,0,sizeof(highways));
    memset(concern,0,sizeof(concern));
    cin>>n;
    cin>>m;
    cin>>k;
    for(i=0;i>c1;
        cin>>c2;
        highways[c1][c2]=true;
        highways[c2][c1]=true;
    }
    for(i=0;i>concern[i];
    }
    for(i=0;i q;
        memset(visit,0,sizeof(visit));
        visit[concern[i]]=1;
        count=0;
        for(j=1;j<=n;j++){
            if(visit[j]!=1){
                queue dq;
                visit[j]=1;
                dq.push(j);
                //對與被占領城市的直接相連的每個城市做BFS
                while(!dq.empty()){
                    tmp=dq.front();
                    dq.pop();
                    for(t=0;t<=n;t++){
                        if(highways[tmp][t]!=false&&visit[t]!=1){
                            dq.push(t);
                            visit[t]=1;
                        }
                    }
                }
                /*如果現有節點可以遍歷所有了所有點，不需要加邊，退出。
                否則獨立“孤島”+1，通過下一個直接相連節點且未被訪問的節點，尋找下一個孤島*/
                count++;
            }
        }
        cout<