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题目背景:
分析:矩阵树定理
求有向图的有根树形图,显然对于有根的树形图是可以直接用矩阵树定理的,那么直接建出入度矩阵和邻接表矩阵,入度矩阵 - 邻接表矩阵获得基尔霍夫矩阵。直接去掉1号点所在行列直接求就可以了。复杂度O(n3)
Source:
/* created by scarlyw */ #include <cstdio> #include <string> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <cctype> #include <vector> #include <set> #include <queue> inline char read() { static const int IN_LEN = 1024 * 1024; static char buf[IN_LEN], *s, *t; if (s == t) { t = (s = buf) + fread(buf, 1, IN_LEN, stdin); if (s == t) return -1; } return *s++; } ///* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = read(), iosig = false; !isdigit(c); c = read()) { if (c == -1) return ; if (c == '-') iosig = true; } for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int OUT_LEN = 1024 * 1024; char obuf[OUT_LEN], *oh = obuf; inline void write_char(char c) { if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf; *oh++ = c; } template<class T> inline void W(T x) { static int buf[30], cnt; if (x == 0) write_char('0'); else { if (x < 0) write_char('-'), x = -x; for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48; while (cnt) write_char(buf[cnt--]); } } inline void flush() { fwrite(obuf, 1, oh - obuf, stdout); } /* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int MAXN = 250 + 10; const int mod = 10007; int n, m, x, y; int a[MAXN][MAXN]; inline int mod_pow(int a, int b) { int ans = 1; for (; b; b >>= 1, a = (long long)a * a % mod) if (b & 1) ans = (long long)ans * a % mod; return ans; } inline void read_in() { R(n), R(m); for (int i = 1; i <= m; ++i) { R(x), R(y), x--, y--; if (x == y) continue ; a[x][x]++, a[y][x]--; } } inline void gauss() { int sign = 1; int ans = 1; for (int i = 1; i < n; ++i) { int pos = -1; for (int j = i; j < n; ++j) if (a[j][i] != 0) { pos = j; break ; } if (pos != i) { sign = -sign; for (int k = i; k < n; ++k) std::swap(a[i][k], a[pos][k]); } int inv = mod_pow(a[i][i], mod - 2); for (int j = i + 1; j < n; ++j) { int t = (long long)a[j][i] * inv % mod; for (int k = i; k < n; ++k) a[j][k] = (a[j][k] - t * a[i][k] % mod + mod) % mod; } ans = ans * a[i][i] % mod; } std::cout << (sign * ans + mod) % mod; } int main() { // freopen("sns.in", "r", stdin); // freopen("sns.out", "w", stdout); read_in(); gauss(); return 0; }