HDU3811 Permutation —— 状压DP

题目链接：https://vjudge.net/problem/HDU-3811

Permutation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 496    Accepted Submission(s): 238

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

 

Author

hanshuai

 

Source

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

 

Recommend

lcy

 

 

题意：

给出m个（A，B），问n的全排列中有多少个满足：至少存在一个i，使得第Ai位为Bi？

 

 

题解：

1.状压DP，设dp[status][has]为：状态为status（前面含有哪几个数），且是否已经满足要求（has）的情况下有多少种。

2.剩下的就是类似TSP的状态转移了（感觉又像是TSP，又像是数位DP）。

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e5;
 const int MAXN = (<<)+;

 bool g[][];
 LL dp[MAXN][];
 int cnt[MAXN];

 void init()
 {
     for(int s = ; s<MAXN; s++)
     {
         cnt[s] = ;
         for(int j = ; j<; j++)
             if(s&(<<j)) cnt[s]++;
     }
 }

 int main()
 {
     init();
     int T, n, m, kase = ;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d", &n, &m);
         memset(g, false, sizeof(g));
         for(int i = ; i<=m; i++)
         {
             int u, v;
             scanf("%d%d", &u, &v);
             g[u][v] = true;
         }

         memset(dp, , sizeof(dp));
         dp[][] = ;
         for(int s = ; s<(<<n); s++)
         {
             for(int i = ; i<; i++)
             {
                 for(int j = ; j<n; j++)
                 if(!(s&(<<j)))
                     dp[s|(<<j)][i|g[cnt[s]+][j+]] += dp[s][i];
             }
         }
         printf("Case %d: %lld\n", ++kase, dp[(<<n)-][]);
     }
 }