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 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-); 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 = ; i <= n; i++) fa[i] = i;
for(int i = ; i <= m; i++) {
scanf("%d%d%d", &a[i], &b[i], &w[i]);
vc[w[i]].push_back(i);
}
for(int i = ; i <= ; 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 = ; i <= num; i++) scanf("%d", &e[i]);
for(int i = ; i <= num; i++) {
if(va[e[i]] == vb[e[i]]) {
flag = false;
break;
}
}
if(!flag) {
puts("NO");
continue;
}
sort(e + , e + + num, cmp);
for(int L = , R = ; 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 ;
} /*
*/