hdu 4408 Minimum Spanning Tree

题目连接：点击打开链接

解法：利用kruskal算法 把图划分成森林， 同一点有相同最小的权值到别的点， 通过determinant计算树的课数。

总结：模板 + 自己不太懂 = 记录 + 重新学

代码君：

#include <stdio.h>
#include <iostream>
#include <vector>

#define LL long long

using namespace std;

const int MAX = 105;
const int MAXE = 1005;

struct node
{
    int set[MAX];
    void init(int n)
    {
        for (int i = 0; i <= n; i++)
            set[i] = i;
    }
    int find(int x)
    {
        if (set[x] != x)
            set[x] = find(set[x]);
        return set[x];
    }
    int Union(int x, int y)
    {
        int xx = find(x);
        int yy = find(y);
        if (xx == yy)
            return -1;
        set[xx] = yy;
        return 1;
    }
};

struct Node
{
    int u, v, dis;
};

node a, b, c;
int n, m;
Node e[MAXE];
int visit[MAX];
vector<int> g[MAX];
LL p[MAX][MAX], MOD, deg[MAX][MAX];

bool cmp(Node a, Node b)
{
    return a.dis < b.dis;
}

LL DET(LL a[][MAX], int n)
{
    LL temp = 1, t;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            a[i][j] %= MOD;
    for (int i = 1; i < n; i++)
    {
        for (int j = i + 1; j < n; j++)
            while (a[j][i])
            {
                t = a[i][i] / a[j][i];
                for (int k = i; k < n; k++)
                {
                    a[i][k] -= a[j][k] * t;
                    a[i][k] %= MOD;
                }
                for (int k = i; k < n; k++)
                {
                    t = a[i][k];
                    a[i][k] = a[j][k];
                    a[j][k] = t;
                }
                temp = -temp;
            }
            temp = temp * a[i][i] % MOD;
    }
    return (temp + MOD) % MOD;
}

LL cal_MST_cout()
{
    sort(e + 1, e + m + 1, cmp);
    int pre = e[1].dis;
    LL ans = 1;
    a.init(n);
    b.init(n);
    memset(visit, false, sizeof(visit));
    memset(deg, false, sizeof(deg));
    for (int i = 0; i <= n; i++)
        g[i].clear();
    for (int t = 1; t <= m + 1; t++)
    {
        if (e[t].dis != pre || t == m + 1)
        {
            for (int i = 1; i <= n; i++)
                if (visit[i])
                {
                    int k = b.find(i);
                    g[k].push_back(i);
                    visit[i] = 0;
                }
            for (int i = 1; i <= n; i++)
                if (g[i].size())
                {
                    memset(p, false, sizeof(p));
                    for (int j = 0; j < g[i].size(); j++)
                        for (int k = j + 1; k < g[i].size(); k++)
                        {
                            int x = g[i][j];
                            int y = g[i][k];
                            p[j][k] = p[k][j] = -deg[x][y];
                            p[j][j] += deg[x][y];
                            p[k][k] += deg[x][y];
                        }
                    ans = ans * DET(p, g[i].size()) % MOD;
                    for (int j = 0; j < g[i].size(); j++)
                        a.set[g[i][j]] = i;
                }
            memset(deg, false, sizeof(deg));
            for (int i = 1; i <= n; i++)
            {
                b.set[i] = a.find(i);
                g[i].clear();
            }
            if (t == m + 1)
                break;
            pre = e[t].dis;
        }
        int x = a.find(e[t].u);
        int y = a.find(e[t].v);
        if (x == y)
            continue;
        visit[x] = visit[y] = 1;
        b.Union(x, y);
        deg[x][y]++;
        deg[y][x]++;
    }
    if (!m)
        return false;
    for (int i = 2; i <= n; i++)
        if (b.find(i) != b.find(1))
            return false;
    return ans;
}

int main()
{
    while (~scanf("%d%d%lld", &n, &m, &MOD) && (n | m | MOD))
    {
        for (int i = 1; i <= m; i++)
            scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].dis);
        printf("%lld\n", cal_MST_cout());
    }
}