hdu 3811

Permutation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 372    Accepted Submission(s): 161

Problem Description

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.

 

Input

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

 

Output

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

 

Sample Input

3 3 2 1 1 2 1 3 2 1 1 2 2 4 3 1 1 1 2 1 3

 

Sample Output

Case 1: 4 Case 2: 3 Case 3: 18

 

 

状态压缩就好 state为某个数取或者不取

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
using namespace std;
long long mp[20][20],A[20],dp[(1<<17)+5];
int n,m;
void AA()
{
    A[0]=1;
    for(int i=1;i<=18;i++)
        A[i]=i*A[i-1];
}
int main()
{
    int tt,ca=1;
    scanf("%d",&tt);
    AA();
    while(tt--)
    {
        memset(mp,0,sizeof(mp));
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            a--,b--;
            mp[a][b]=1;
        }
        dp[0]=1;
        for(int i=0;i<n;i++)
        {
            for(int j=(1<<n)-1;j>=0;j--)
            {
                if(dp[j]==0)
                    continue;
                for(int k=0;k<n;k++)
                {
                    if((j&(1<<k)))
                        continue;
                    if(mp[i][k])
                        continue;
                    dp[j|(1<<k)]+=dp[j];
                }
            }
        }
        printf("Case %d: %I64d
",ca++,A[n]-dp[(1<<n)-1]);
    }
    return 0;
}