POJ 2441 Arrange the Bulls (状压DP)

题意：n头牛，m个位置，每头牛有各自喜欢的位置，问安排这n头牛使得每头牛都在各自喜欢的位置有几种安排方法。

析：dp[i][s] 表示前 i 头牛，已经占的位置是 s，有多少种安排方法，其他的就很简单了，注意用滚动数组　　。

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>
#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 = 1e6 + 10;
const int mod = 100000000;
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;
}
int dp[2][1<<20];
int a[30][30];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 1; i <= n; ++i){
      scanf("%d", &a[i][0]);
      for(int j = 1; j <= a[i][0]; ++j){
        scanf("%d", &a[i][j]);
        --a[i][j];
      }
    }
    memset(dp[0], 0, sizeof dp[0]);
    dp[0][0] = 1;
    int cnt = 1;
    int all = 1 << m;
    for(int i = 1; i <= n; ++i, cnt ^= 1){
      memset(dp[cnt], 0, sizeof dp[cnt]);
      for(int j = 0; j < all; ++j){
        if(!dp[cnt^1][j])  continue;
        for(int k = 1; k <= a[i][0]; ++k){
          if(j&(1<<a[i][k]))  continue;
          dp[cnt][j|(1<<a[i][k])] += dp[cnt^1][j];
        }
      }
    }
    int ans = 0;
    for(int i = 0; i < all; ++i)
      ans += dp[cnt^1][i];
    printf("%d
", ans);
  }
  return 0;
}