hdu 3811 Permutation

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#permutation #output #each #integer #bi #input

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

Sample Output

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

Author

hanshuai

Source

The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Preliminary

又是一题状态dp……当年比赛的时候RE到死没出来，今天1Y……rp咩？

#include <stdio.h>
#include <string.h>

int map[20][20];
long long dp[1<<18];
long long fact[18];

int main()
{
    int i,j,n,T,m,x,y,p,k,cnt;
    scanf("%d",&T);
    fact[0]=1;
    for (i=1;i<18;i++)
    {
        fact[i]=fact[i-1]*i;
    }
    cnt=1;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        for (i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            x--;
            y--;
            map[x][y]=1;
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for (i=0;i<n;i++)
        {
            for (j=(1<<n)-1;j>=0;j--)
            {
                if (dp[j]==0) continue;
                for (k=0;k<n;k++)
                {
                    if ((j & (1<<k))!=0) continue;
                    if (map[i][k]==1) continue;
                    p=(j | (1<<k));
                    dp[p]+=dp[j];
                }
            }
        }
        printf("Case %d: %I64d\n",cnt++,fact[n]-dp[(1<<n)-1]);
    }
    return 0;
}