1013. Battle Over Cities (25)

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本文介绍了一种算法，用于计算当某个城市被敌方占领后，为了保持剩余城市的连接性，需要修复多少条高速公路。通过输入城市数量、剩余高速公路数量及关注的城市列表，算法能够快速给出修复方案。

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

Sample Output

1
0
0

参考代码：

#include <iostream>
#include <string.h>
 using namespace std;
 int roads[1000][1000];
 int visited[1000];
 int N,M,K;
 void DFS(int t){
    visited[t] = 1;
    for(int i=0;i<N;i++){
        if(!visited[i] && roads[t][i]==1){
            DFS(i);
        }
    }
 }

 int main(){
    cin>>N>>M>>K;
    for(int i=0;i<M;i++){
        int t_a,t_b;
        cin>>t_a>>t_b;
        roads[t_a-1][t_b-1] = 1;
        roads[t_b-1][t_a-1] = 1;
    }
    for(int i=0;i<K;i++){
        int lostCity;
        cin>>lostCity;
        memset(visited,0,sizeof(visited));
        visited[lostCity-1] = 1;   //表示这个城市已经丢失，不用再遍历
        int num = 0;    //记录图中独立子图的个数
        for(int j=0;j<N;j++){
            if(visited[j]==0){
                DFS(j);
                num++;
            }
        }
        cout<<num-1<<endl;
    }
}