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
", ans);
  }
  return 0;
}