HDU 3811 Permutation 状压dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3811

Permutation

Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
#### 问题描述
> In combinatorics a permutation of a set S with N elements is a listing of the elements of S in some order (each element occurring exactly once). There are N! permutations of a set which has N elements. For example, there are six permutations of the set {1,2,3}, namely [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
> But Bob think that some permutations are more beautiful than others. Bob write some pairs of integers(Ai, Bi) to distinguish beautiful permutations from ordinary ones. A permutation is considered beautiful if and only if for some i the Ai-th element of it is Bi. We want to know how many permutations of set {1, 2, ...., N} are beautiful.

输入

The first line contains an integer T indicating the number of test cases.

There are two integers N and M in the first line of each test case. M lines follow, the i-th line contains two integers Ai and Bi.

Technical Specification

1. 1 <= T <= 50
2. 1 <= N <= 17
3. 1 <= M <= N*N
4. 1 <= Ai, Bi <= N

输出

For each test case, output the case number first. Then output the number of beautiful permutations in a line.

样例输入

3

3 2

1 1

2 1

3 2

1 1

2 2

4 3

1 1

1 2

1 3

样例输出

Case 1: 4

Case 2: 3

Case 3: 18

题意

给你一些条件：(a,b)，在a这个位置要放b，求至少满足其中一个条件的1到n的排列有多少种。

题解

首先，从正面很明显可以看出是容斥，考虑条件间的容斥，但是发现条件有17*17个，因此不可行。

然后可以从反面出发，要求至少满足一个就是求所有的-一个都不满足的，所以我们只要求出一个都不满足的就可以了。

这个可以用dp做，dp[i][j]表示前i位放的状态为j的情况。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=17;

LL dp[maxn][1<<maxn];
int n,m;

int mp[maxn][maxn];

int main() {
    int tc,kase=0;
    scf("%d",&tc);
    while(tc--){
        scf("%d%d",&n,&m);
        clr(mp,0);
        rep(i,0,m){
            int u,v;
            scf("%d%d",&u,&v); u--,v--;
            mp[u][v]=true;
        }

            ///初始化
        clr(dp,0);
        for(int i=0;i<n;i++){
            if(mp[0][i]==0){
                dp[0][1<<i]=1;
            }
        }

            ///递推
        for(int i=1;i<n;i++){
            for(int j=0;j<(1<<n);j++){
                for(int k=0;k<n;k++){
                    if(!(j&(1<<k))&&!mp[i][k]){
                        dp[i][j^(1<<k)]+=dp[i-1][j];
                    }
                }
            }
        }

        LL ans=1;
        for(int i=2;i<=n;i++) ans*=i;
        ans-=dp[n-1][(1<<n)-1];
        prf("Case %d: %lld\n",++kase,ans);
    }
    return 0;
}

//end-----------------------------------------------------------------------