[BZOJ1543] 生成树计数 (Kruskal)

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 struct edge
 4 {
 5     int u, v, w, x;
 6     inline bool operator< (const edge &rhs) const
 7     {
 8         return x < rhs.x;
 9     }
10 }e[100005];
11 struct count
12 {
13     int l, r, use;
14 }g[100005];
15 int n, m, fa[50005], siz[50005];
16    
17 int getfa(int x)
18 {
19     return fa[x] == x ? x : getfa(fa[x]);
20 }
21    
22 void link(int u, int v)
23 {
24     if(siz[u] > siz[v]) fa[v] = u, siz[u] += siz[v];
25     else fa[u] = v, siz[v] += siz[u];
26 }
27    
28 bool Kruskal()
29 {
30     int cnt = 0, u, v;
31     for(int i = 1; i <= m; ++i)
32     {
33         u = getfa(e[i].u), v = getfa(e[i].v);
34         if(u != v)
35         {
36             link(u, v);
37             ++g[e[i].w].use;
38             if(++cnt == n - 1) return true;
39         }
40     }
41     return false;
42 }
43    
44 int DFS(int w, int i, int k)
45 {
46     if(k == g[w].use) return 1;
47     if(i > g[w].r) return 0;
48     int ans = 0, u = getfa(e[i].u), v = getfa(e[i].v);
49     if(u != v)
50     {
51         link(u, v);
52         ans = DFS(w, i + 1, k + 1);
53         fa[u] = u, fa[v] = v;
54     }
55     return ans + DFS(w, i + 1, k);
56 }
57    
58 int main()
59 {
60     int u, v, w, ans;
61     cin >> n >> m;
62     for(int i = 1; i <= n; ++i)
63         fa[i] = i, siz[i] = 1;
64     for(int i = 1; i <= m; ++i)
65     {
66         cin >> u >> v >> w;
67         e[i] = (edge){u, v, 0, w};
68     }
69     sort(e + 1, e + m + 1);
70     w = 0;
71     for(int i = 1; i <= m; ++i)
72         if(e[i].x == e[i - 1].x) e[i].w = w;
73         else
74         {
75             g[w].r = i - 1;
76             e[i].w = ++w;
77             g[w].l = i;
78         }
79     g[w].r = m;
80     ans = Kruskal();
81     for(int i = 1; i <= n; ++i)
82         fa[i] = i, siz[i] = 1;
83     for(int i = 1; i <= w; ++i)
84     {
85         ans = ans * DFS(i, g[i].l, 0) % 1000003;
86         for(int j = g[i].l; j <= g[i].r; ++j)
87         {
88             u = getfa(e[j].u), v = getfa(e[j].v);
89             if(u != v) link(u, v);
90         }
91     }
92     cout << ans << endl;
93     return 0;
94 }