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\n", 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