PAT 1013 Battle Over Cities

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#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;}

通過dfs把可以聯通的地區的節點設定為visited=true，這樣對visited=false的節點進行了幾次dfs就有幾個不相連的地區，除去一個被佔領的city，剩下的就是需要重新串連起來的地區，地區個數-1就是需要修複的路的數量

PAT 1013 Battle Over Cities