[JSOI2008]巨额奖金（最小生成树计数）

基于kruscal的贪心 可以有如下推论：两个不同的最小生成树所用的的相同边权的边的数量相同。

可以做这样的理解：当进行到选第K大的边时，联通快数是一定的所以的、一定需要K- 1个该权值边 （不知道对不对 欢迎TuneProverbs指正）

SB没想好怎么存水体写半天。。。。。。。

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1500;typedef long long LL;
vector<int> v[N];
vector<int>::iterator it, ptr;
int n, m, p[N], cnt[N], cc, r[N];
LL ans;
bool vis[N], cantvisit[N], noans;
struct Q {
    int x, y, w, num;
    inline bool operator < (const Q & a) const {
        return w < a.w;
    }
}e[N];
inline int F (int x) { return p[x] == x ? x : p[x] = F (p[x]); }
void ksc ()
{
    for (int i = 1; i <= n; i ++)
        p[i] = i;
    sort (e + 1, e + 1 + m);
    for (int i = 1; i <= m; i ++)
        if (F (e[i].x) != F (e[i].y))
        {
            p[F (e[i].x)] = F (e[i].y);
            ans += e[i].w;
            int j (0);
            for (int j = 1; j <= cc; j ++)
                if (r[j] == e[i].w)
                    cnt[j] ++;
        }
    for (int i = 2; i <= n; i ++)
        if (F (i) != F (i - 1))
            noans = true;
    //for (int i = 1; i <= 5; i ++)
    //    printf ("%d\n", cnt[i]);
    //printf ("%d\n", ans);
}
bool ck ()
{
    for (int i = 1; i <= n; i ++)
        p[i] = i;
    sort (e + 1, e + 1 + m);
    LL z (0);
    for (int i = 1; i <= m; i ++)
    {
        if (cantvisit[e[i].num]) continue;
        if (F (e[i].x) != F (e[i].y))
        {
            p[F (e[i].x)] = F (e[i].y);
            z += e[i].w;
        }
    }
    //printf ("%d\n", z);
    return z == ans;
}
inline int calc (int j)
{
    int z (0);
    for (; j; j -= j & -j) z ++;
    return z;
}
int main ()
{
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++)
        scanf ("%d%d%d", &e[i].x, &e[i].y, &e[i].w), e[i].num = i;
    for (int i = 1; i <= m; i ++)
    {
        if (vis[i]) continue;
        cc ++;
        r[cc] = e[i].w;
        v[cc].push_back (i);
        for (int j = i + 1; j <= m; j ++)
            if (e[j].w == e[i].w)
            {
                vis[j] = true;
                v[cc].push_back (j);
            }
    }
    ksc ();
    if (noans == true)
    {
        puts ("0");
        return 0;
    }
    int z (1);
    for (int j = 1; j <= cc; j ++)
    {
        int tmp (0);
        //printf ("*%d\n", v[j].size ());
        for (int i = 0; i < (1 << v[j].size ()); i ++)
        {
            if (calc (i) == cnt[j])
            {
                int k (0);
                for (ptr = v[j].begin (); ptr != v[j].end (); ptr ++, k ++)
                    if ((i >> k) & 1)
                        ;
                    else
                        cantvisit[*ptr] = true;
                //for (int i = 1; i <= m; i ++)
                //    printf ("%d", cantvisit[i]);puts ("");
                int t (0);
                if (t = ck ())
                    tmp ++;
                memset (cantvisit, 0, sizeof cantvisit);
            }
        }
        //printf ("%d\n", tmp);
        if (tmp != 0)
            z = ((long long)z * tmp) % 31011;
    }
    printf ("%d\n", z);
    return 0;
}