PAT:1013 Battle Over Cities (25分)(DFS)

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本文探讨了在战争背景下，确保城市间高速公路连通性的算法挑战。通过输入城市数量、剩余高速公路及关注城市，快速计算若某城市被敌方占领，需修复的高速公路数量，以维持其余城市的连接。示例展示了算法的应用过程。

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 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:

Sample Output:

1
0
0

对于图来讲，如果DFS后没有让状态全部改变，那么就说明有地方时不连通的，所以至少得建一个桥，建完之后再DFS，如果还不能让全部状态改变，说明还得建桥，这样等到全部状态都改变了，也说明桥建完了

#include<bits/stdc++.h>
using namespace std;
vector<vector<int>>v;
bool visit[10010]={false};
void dfs(int index)
{
    visit[index]=true;
    for(int i=0;i<v[index].size();i++){
        if(visit[v[index][i]]==false)
            dfs(v[index][i]);
    }
}

int main()
{
    int n,m,k;
    scanf("%d %d %d",&n,&m,&k);
    v.resize(n+1);
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d %d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    for(int i=0;i<k;i++){
        fill(visit,visit+10010,false);
        int sum=-1;
        int a;
        scanf("%d",&a);
        visit[a]=true;
        for(int j=1;j<=n;j++){
            if(visit[j]==false)
            {
             dfs(j);
             sum++;
            }
        }
        cout<<sum<<endl;

    }

}