【51Nod1833】环

【题目链接】

点击打开链接

【思路要点】

原题实际上等价于计算二分图不同的完美匹配的数量。

将每个点拆成入点和出点，每条边由出点连向入点，形成的二分图的完美匹配和用简单环覆盖原图的方案一一对应。

状压 $DP$ 即可。

时间复杂度 $O(2^N*N)$ 。

【代码】


#include<bits/stdc++.h>

using namespace std;
const int MAXN = 22;
const int MAXS = 1 << 20;
const int P = 998244353;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
int n, m, dp[MAXN][MAXS], bit[MAXN], bits[MAXS];
vector <int> a[MAXN];
void update(int &x, int y) {
  x += y;
  if (x >= P) x -= P;
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= m; i++) {
      int x, y; read(x), read(y);
      a[x].push_back(y);
  }
  for (int i = 1; i <= n; i++)
      bit[i] = 1 << (i - 1);
  int goal = (1 << n) - 1;
  for (int i = 1; i <= goal; i++)
      bits[i] = bits[i - (i & -i)] + 1;
  dp[0][0] = 1;
  for (int i = 0; i <= n - 1; i++)
  for (int s = 0; s <= goal; s++) {
      if (bits[s] != i) continue;
      for (unsigned j = 0; j < a[i + 1].size(); j++) {
          int tmp = bit[a[i + 1][j]];
          if ((s & tmp) == 0) update(dp[i + 1][s | tmp], dp[i][s]);
      }
  }
  writeln(dp[n][goal]);
  return 0;
}

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