感觉这种最小生成树上的啥题都差不多的解法。。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); int n, m, q, a[N], b[N], w[N], va[N], vb[N], e[N]; vector<int> vc[N]; int fa[N]; int getRoot(int x) { return fa[x] == x ? x : fa[x] = getRoot(fa[x]); } bool cmp(const int& a, const int& b) { return w[a] < w[b]; } int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) fa[i] = i; for(int i = 1; i <= m; i++) { scanf("%d%d%d", &a[i], &b[i], &w[i]); vc[w[i]].push_back(i); } for(int i = 1; i <= 500000; i++) { if(!SZ(vc[i])) continue; for(auto& x : vc[i]) { va[x] = getRoot(a[x]); vb[x] = getRoot(b[x]); } for(auto& x : vc[i]) { fa[getRoot(b[x])] = getRoot(a[x]); } } scanf("%d", &q); while(q--) { bool flag = true; int num; scanf("%d", &num); for(int i = 1; i <= num; i++) scanf("%d", &e[i]); for(int i = 1; i <= num; i++) { if(va[e[i]] == vb[e[i]]) { flag = false; break; } } if(!flag) { puts("NO"); continue; } sort(e + 1, e + 1 + num, cmp); for(int L = 1, R = 1; L <= num && flag; L = R) { while(R <= num && w[e[R]] == w[e[L]]) R++; for(int i = L; i < R; i++) { int x = va[e[i]], y = vb[e[i]]; fa[x] = x, fa[y] = y; } for(int i = L; i < R; i++) { int x = va[e[i]], y = vb[e[i]]; int u = getRoot(x), v = getRoot(y); if(u == v) { flag = false; break; } fa[v] = u; } } if(flag) puts("YES"); else puts("NO"); } return 0; } /* */