dlutoj 1081 Prufer 序列

題目鏈接

分析：這道題並不難。看懂了prufer序列就能明白，序列中出現的節點應該是一顆生成樹的非葉子節點，而且題目已經降低了難度，序列中的相同節點只計算一次。所以，題目轉變成了統計每個節點作爲非葉子節點的生成樹的個數。對一張圖，統計生成樹的個數要用到基爾霍夫矩陣（矩陣樹定理），這個不明白的話先要學習一個。統計非葉子節點的想法很簡單，把這個點先移除，剩餘的點統計生成樹的個數，那麼該節點作爲葉子節點的生成樹個數就可以由節點的度乘上剩餘生成樹個數計算得出。然後用總的生成樹個數減去就可以算出來。剩下的問題就變成了矩陣行列式的計算問題了。線代沒學好＝ ＝忘記在交換的時候乘-1了。

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代碼：

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)
#define   DEBUG         freopen("#.txt", "r", stdin)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   ll            long long
#define   ull           unsigned long long
#define   lowbit(x)     (x&-x)

const int mod = 1e9 + 7;
const int maxn = 105;

long long d[maxn][maxn], g[maxn][maxn];
long long a[maxn][maxn], t[maxn][maxn];
long long deg[maxn], tt;
int N, M;

void Debug(int sz) {
    cout<<endl;
    for (int i = 0; i < sz; i ++) {
        for (int j = 0; j < sz; j ++)
            cout<<t[i][j]<<" ";
        cout<<endl;
    }
    cout<<endl;
}

long long det(int sz) {
    long long ans = 1;
    bool sign = 0;
    for (int i = 0; i < sz; i ++)
        for (int j = 0; j < sz; j ++)
            t[i][j] = (t[i][j] % mod + mod) % mod;
    for (int i = 0; i < sz; i ++) {
        for (int j = i + 1; j < sz; j ++) {
            int x = i, y = j;
            while (t[y][i]) {
                ll tmp = t[x][i] / t[y][i];
                for (int k = i; k < sz; k ++)
                    t[x][k] = ((t[x][k] - t[y][k] * tmp) % mod + mod) % mod;
                swap(x, y);
            }
            if (x != i) {
                for (int k = 0; k < sz; k ++)
                    swap(t[i][k], t[x][k]);
                sign ^= 1;
            }
        }
        if (!t[i][i]) return 0;
        ans = ans * t[i][i] % mod;
    }
    if (sign) ans = (mod - ans) % mod;
    return ans;
}

int main(int argc, char const *argv[]) {
    int T; cin>>T;
    while (T --) {
        cin>>N>>M;
        memset(g, 0, sizeof(g));
        memset(t, 0, sizeof(t));
        memset(deg, 0, sizeof(deg));

        for (int i = 0; i < M; i ++) {
            int u, v;
            scanf("%d%d", &u, &v); u --, v --;
            g[u][v] = g[v][u] = 1;
            deg[u] ++, deg[v] ++;
        }
        for (int i = 0; i < N; i ++)
            d[i][i] = deg[i];

        for (int i = 0; i < N; i ++)
            for (int j = 0; j < N; j ++)
                t[i][j] = a[i][j] = d[i][j] - g[i][j];

        long long ans = 0;
        long long num = det(N - 1);

        if (num < 0) {
            cout<<0<<endl;
            continue;
        }

        for (int k = 0; k < N; k ++) {
            memset(t, 0, sizeof(t));

            for (int i = 0; i < k; i ++) {
                for (int j = 0; j < k; j ++)
                    if (i != j && g[i][j]) {
                        t[i][j] = -1;
                        t[i][i] ++;
                    }
                for (int j = k + 1; j < N; j ++)
                    if (i != j && g[i][j]) {
                        t[i][j - 1] = -1;
                        t[i][i] ++;
                    }
            }
            for (int i = k + 1; i < N; i ++) {
                for (int j = 0; j < k; j ++)
                    if (i != j && g[i][j]) {
                        t[i - 1][j] = -1;
                        t[i - 1][i - 1] ++;
                    }
                for (int j = k + 1; j < N; j ++)
                    if (i != j && g[i][j]) {
                        t[i - 1][j - 1] = -1;
                        t[i - 1][i - 1] ++;
                    }
            }

            // Debug(N - 1);
            long long res = det(N - 2);
            if (res < 0) continue;
            res = ((num - res * deg[k]) % mod + mod) % mod;
            ans = ((ans + res * (k + 1)) % mod + mod) % mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}