UVa 10766 Organising the Organisation (生成树计数)

题意：给定一个公司的人数，然后还有一个boss，然后再给定一些人，他们不能成为直属上下级关系，问你有多少种安排方式（树）。

析：就是一个生成树计数，由于有些人不能成为上下级关系，也就是说他们之间没有边，没说的就是有边，用Matrix-Tree定理，很容易就能得到答案，注意题目给定的可能有重复的。

对于基尔霍夫矩阵，就是度数矩阵，减去邻接矩阵，一处理就OK了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <bitset>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 100 + 50;

const int mod = 500500;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r > 0 && r <= n && c > 0 && c <= m;

}

LL a[maxn][maxn];

int in[maxn];

LL solve(){

  LL ans = 1;

  for(int i = 1; i < n; ++i){

    for(int j = 1+i; j < n; ++j)

      while(a[j][i]){

        LL t = a[i][i] / a[j][i];

        for(int k = i; k < n; ++k)

          a[i][k] -= t * a[j][k];

        for(int k = i; k < n; ++k)

          swap(a[i][k], a[j][k]);

      }

    if(a[i][i] == 0)  return 0;

    ans *= a[i][i];

  }

  return abs(ans);

}

int main(){

  int k;

  while(scanf("%d %d %d", &n, &m, &k) == 3){

    for(int i = 1; i <= n; ++i){

      for(int j = i+1; j <= n; ++j)

        a[i][j] = a[j][i] = -1;

      in[i] = n - 1;

    }

    for(int i = 0; i < m; ++i){

      int u, v;

      scanf("%d %d", &u, &v);

      if(a[u][v] == -1)  --in[v],  --in[u];

      a[u][v] = a[v][u] = 0;

    }

    for(int i = 1; i <= n; ++i)  a[i][i] = in[i];

    LL ans = solve();

    printf("%lld\n", ans);

  }

  return 0;

}