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-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e5;
const int MAXN = (1<<17)+10;

bool g[20][20];
LL dp[MAXN][2];
int cnt[MAXN];

void init()
{
    for(int s = 0; s<MAXN; s++)
    {
        cnt[s] = 0;
        for(int j = 0; j<17; j++)
            if(s&(1<<j)) cnt[s]++;
    }
}

int main()
{
    init();
    int T, n, m, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        memset(g, false, sizeof(g));
        for(int i = 1; i<=m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            g[u][v] = true;
        }

        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int s = 0; s<(1<<n); s++)
        {
            for(int i = 0; i<2; i++)
            {
                for(int j = 0; j<n; j++)
                if(!(s&(1<<j)))
                    dp[s|(1<<j)][i|g[cnt[s]+1][j+1]] += dp[s][i];
            }
        }
        printf("Case %d: %lld
", ++kase, dp[(1<<n)-1][1]);
    }
}