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
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.
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
3 2
1 1
2 1
3 2
1 1
2 2
4 3
1 1
1 2
1 3
Case 2: 3
Case 3: 18
- #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)-][]);
- }
- }
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