题目链接
https://www.patest.cn/contests/gplt/L2-025
思路 只要把被攻下的城市标记一下 与 其他城市之间的通路都取消
然后判断一下剩下的城市 是否都是孤立的 就可以
AC代码
#include <cstdio> #include <cstring> #include <ctype.h> #include <cstdlib> #include <cmath> #include <climits> #include <ctime> #include <iostream> #include <algorithm> #include <deque> #include <vector> #include <queue> #include <string> #include <map> #include <stack> #include <set> #include <numeric> #include <sstream> #include <iomanip> #include <limits> #define CLR(a) memset(a, 0, sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; typedef pair <int, int> pii; typedef pair <ll, ll> pll; const double PI = acos(-1); const double E = exp(1); const double eps = 1e-6; const int INF = 0x3f3f3f3f; const int maxn = 1e4 + 5; const int MOD = 1e9 + 7; vector <int> G[maxn]; int v[maxn]; void init() { CLR(G); } int main() { int n, m; scanf("%d%d", &n, &m); int x, y; for (int i = 0; i < m; i++) { scanf("%d%d", &x, &y); G[x].pb(y); G[y].pb(x); } int T; scanf("%d", &T); while (T--) { int k; scanf("%d", &k); map <int, int> M; for (int j = 0; j < k; j++) { int num; scanf("%d", &num); M[num] = 1; } int flag = 1; vector <int>::iterator it; for (int i = 0; i < n; i++) { if (M[i]) continue; for (it = G[i].begin(); it != G[i].end(); it++) { if (M[*it] == 0) { flag = 0; break; } } } printf("%s\n", flag? "YES":"NO"); } }转载于:https://www.cnblogs.com/Dup4/p/9433148.html
