【BZOJ1016】【Luogu P4208】 [JSOI2008]最小生成树计数 最小生成树,矩阵树定理

蛮不错的一道题,遗憾就遗憾在数据范围会导致暴力轻松跑过。

最小生成树的两个性质:

  • 不同的最小生成树,相同权值使用的边数一定相同。

  • 不同的最小生成树,将其都去掉同一个权值的所有边,其连通性一致。

这样我们随便跑一个\(MST\),就可以知道所有\(MST\)边的构造情况。由于性质二,我们可以考虑枚举每一种权值的所有边,保留所有非此权值的树边,看可以连出来多少种不同的最小生成树。也就是按照权值构造最小生成树,这个过程满足乘法原理。

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 100 + 5;
const int M = 1000 + 5;
const int Mod = 31011;

struct Len {
    int u, v, w;
    
    bool operator < (Len rhs) const {
        return w < rhs.w;
    }
}l[M];

vector <Len> v;

int n, m, S[N];

int find (int x) {
    return S[x] == x ? x : S[x] = find (S[x]);
}

vector <int> val, use, tot;

vector <int> :: iterator it;

void kruskal () {
    sort (l + 1, l + 1 + m);
    for (int i = 0; i <= n; ++i) S[i] = i;
    for (int i = 1; i <= m; ++i) {
        int fu = find (l[i].u);
        int fv = find (l[i].v);
        it = lower_bound (val.begin (), val.end (), l[i].w); 
        if (it == val.end ()) {
            val.push_back (l[i].w);
            use.push_back (0);
            tot.push_back (1);
        } else {
            tot[it - val.begin ()]++;
        }
        if (fu != fv) {
            S[fu] = fv;
            it = lower_bound (val.begin (), val.end (), l[i].w); 
            use[it - val.begin ()]++;
            v.push_back (l[i]);
        }
    }
}

int mat[N][N];

int gauss (int n) {
    int ret = 1;
    for (int i = 1; i <= n; ++i) {
        for (int k = i + 1; k <= n; ++k) {
            while (mat[k][i]) {
                int d = mat[i][i] / mat[k][i];
                for (int j = i; j <= n; ++j) {
                    (((mat[i][j] -= d * mat[k][j]) %= Mod) += Mod) %= Mod;
                }
                swap (mat[i], mat[k]); ret = -ret;
            }
        }
        (((ret *= mat[i][i]) %= Mod) += Mod) %= Mod;
    }
    return abs (ret);
}

void add_edge (int u, int v) {
    mat[u][u]++; 
    mat[v][v]++;
    mat[u][v]--;
    mat[v][u]--;
}

int sep[N]; 

int solve () {
    kruskal ();
    if (v.size () < n - 1) return 0;
    int ans = 1;
    for (int i = 0; i < val.size (); ++i) {
        memset (mat, 0, sizeof (mat));
        if (use[i] == 0 || tot[i] == use[i]) continue;
        for (int j = 0; j <= n; ++j) S[j] = j;
        for (int j = 0; j < v.size (); ++j) {
            if (v[j].w != val[i]) {
                S[find (v[j].u)] = find (v[j].v);
            }
        }
        int cnt = 0;
        for (int i = 1; i <= n; ++i) {
            sep[++cnt] = find (i);
        }
        sort (sep + 1, sep + 1 + cnt);
        cnt = unique (sep + 1, sep + 1 + cnt) - sep - 1;
        for (int j = 1; j <= m; ++j) {
            if (l[j].w == val[i]) {
                int fu = find (l[j].u);
                int fv = find (l[j].v);
                fu = lower_bound (sep + 1, sep + 1 + cnt, fu) - sep;
                fv = lower_bound (sep + 1, sep + 1 + cnt, fv) - sep;
                add_edge (fu, fv);
            }
        }
        (ans *= gauss (use[i])) %= Mod;
    }
    return ans;
}

signed main () {
//  freopen ("data.in", "r", stdin);
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        static int u, v, w;
        cin >> u >> v >> w;
        l[i] = (Len) {u, v, w};
    }
    cout << solve () << endl;
}
posted @ 2019-05-29 20:32  maomao9173  阅读(...)  评论(... 编辑 收藏