PAT 1013 Battle Over Cities

#include <cstdio>
#include <cstdlib>
#include <vector>

using namespace std;

class City {
public:
    vector<int> adj;
    bool  visited;
    City() : visited(false) {}
};

void dfs(int idx, vector<City>& cities) {
    if (cities[idx].visited) {
        return;
    }

    City& city = cities[idx];
    city.visited = true;
    
    for (int i=0; i<city.adj.size(); i++) {
        dfs(city.adj[i], cities);
    }
}

void reset_cities(vector<City>& cities) {
    int len = cities.size();
    for (int i=0; i<len; i++) {
        cities[i].visited = false;
    }
}

int main() {
    int N = 0, M = 0, K = 0;
    
    scanf("%d%d%d", &N, &M, &K);
    
    vector<City> cities(N + 1);
    
    for (int i=0; i<M; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        cities[a].adj.push_back(b);    
        cities[b].adj.push_back(a);
    }
    
    for (int i=0; i<K; i++) {
        int parts = 0;
        reset_cities(cities);
        int oc = -1;
        scanf("%d", &oc);
        cities[oc].visited = true;
        for (int i=1; i<=N; i++) {
            if (cities[i].visited) continue;
            parts++;
            dfs(i, cities);
        }
        printf("%d
", parts - 1);
    }
    return 0;
}

通过dfs把可以联通的区域的节点设置为visited=true，这样对visited=false的节点进行了几次dfs就有几个不相连的区域，除去一个被占领的city，剩下的就是需要重新连接起来的区域，区域个数-1就是需要修复的路的数量