题面
传送门
Sol
最小生成树的性质：
- 对于每一个\(MST\)，每一种边权所使用的边数相同
- 所有\(MST\)中边权\(≤w\)的边组成的图的连通性相同
那么这道题就枚举没个权值选那些边，如果连的个数和原来的相同就统计
最后乘法原理即可
如果同边权过多就只能用矩阵树定理了
然而我太菜了不会。。
# include <bits/stdc++.h># define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)using namespace std;typedef long long ll;const int Zsy(31011);const int _(105);const int __(1005);IL int Input(){ ???RG int x = 0, z = 1; RG char c = getchar(); ???for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; ???for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); ???return x * z;}int n, m, fa[_], ans = 1, o[__], len, l[__], r[__], id[__], cnt[__], tmp[_];struct Edge{ ???int u, v, w; ???IL bool operator <(RG Edge B) const{ ???????return w < B.w; ???}} edge[__];IL int Find(RG int x){ ???return x == fa[x] ? x : fa[x] = Find(fa[x]);}IL int Count(RG int x){ ???RG int ret = 0; ???for(; x; x -= x & -x) ++ret; ???return ret;}int main(RG int argc, RG char* argv[]){ ???n = Input(), m = Input(); ???for(RG int i = 1; i <= n; ++i) fa[i] = i; ???for(RG int i = 1; i <= m; ++i){ ???????edge[i] = (Edge){Input(), Input(), Input()}; ???????o[i] = edge[i].w, l[i] = m, r[i] = 1; ???} ???sort(edge + 1, edge + m + 1); ???sort(o + 1, o + m + 1), len = unique(o + 1, o + m + 1) - o - 1; ???RG int gg = 0; ???for(RG int i = 1; i <= m; ++i){ ???????RG int fx = Find(edge[i].u), fy = Find(edge[i].v); ???????id[i] = lower_bound(o + 1, o + len + 1, edge[i].w) - o; ???????l[id[i]] = min(l[id[i]], i), r[id[i]] = max(r[id[i]], i); ???????if(Find(fx) == Find(fy)) continue; ???????fa[fx] = fy, ++cnt[id[i]], ++gg; ???} ???if(gg != n - 1) return puts("0"), 0; ???for(RG int i = 1; i <= n; ++i) fa[i] = i; ???for(RG int i = 1; i <= len; ++i){ ???????RG int tot = 0; ???????for(RG int j = 1; j <= n; ++j) tmp[j] = fa[j]; ???????for(RG int j = 0, S = 1 << (r[i] - l[i] + 1); j < S; ++j){ ???????????if(Count(j) != cnt[i]) continue; ???????????RG int flg = 0; ???????????for(RG int k = 1; k <= n; ++k) fa[k] = tmp[k]; ???????????for(RG int k = 0; k <= r[i] - l[i]; ++k) ???????????????if((1 << k) & j){ ???????????????????RG int fx = Find(edge[k + l[i]].u), fy = Find(edge[k + l[i]].v); ???????????????????if(fx == fy){ ???????????????????????flg = 1; ???????????????????????break; ???????????????????} ???????????????????fa[fx] = fy; ???????????????} ???????????if(!flg) ++tot; ???????} ???????ans = ans * tot % Zsy; ???????for(RG int j = 1; j <= n; ++j) fa[j] = tmp[j]; ???????for(RG int j = l[i]; j <= r[i]; ++j){ ???????????RG int fx = Find(edge[j].u), fy = Find(edge[j].v); ???????????if(fx != fy) fa[fx] = fy; ???????} ???} ???printf("%d\n", ans); ???return 0;}