2015 Multi-University Training Contest 10 hdu 5411 CRB and Puzzle

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原文链接：http://www.cnblogs.com/crackpotisback/p/4749113.html

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#java #数据结构与算法

本文详细介绍了CRB使用无限供应的拼图进行拼接的方法，目标是在不超过M次移动的情况下，计算可以组装的不同图案数量，并通过矩阵快速幂和动态规划解决此问题。

CRB and Puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 301 Accepted Submission(s): 127

Problem Description
CRB is now playing Jigsaw Puzzle.
There are N kinds of pieces with infinite supply.
He can assemble one piece to the right side of the previously assembled one.
For each kind of pieces, only restricted kinds can be assembled with.
How many different patterns he can assemble with at most M pieces? (Two patterns P and Q are considered different if their lengths are different or there exists an integer j such that j-th piece of P is different from corresponding piece of Q.)

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers N, M denoting the number of kinds of pieces and the maximum number of moves.
Then N lines follow. i-th line is described as following format.
$k a_1 a_2 ... a_k$
Here k is the number of kinds which can be assembled to the right of the i-th kind. Next k integers represent each of them.
$1 \leq T \leq 20$
$1 \leq N \leq 50$
$1 \leq M \leq 10^5$
$0 \leq k \leq N$
$1 \leq a1 < a2 < … < ak \leq N$

Output
For each test case, output a single integer - number of different patterns modulo 2015.

Sample Input
1
3 2
1 2
1 3
0

Sample Output
6

Hint
possible patterns are ∅, 1, 2, 3, 1→2, 2→3

Author
KUT（DPRK）

解题：矩阵快速幂+动态规划

$$\begin{bmatrix} ret & dp1 & dp2 & dp3 \\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ \end{bmatrix} $$

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 60;
 4 const int mod = 2015;
 5 int n,m;
 6 struct Matrix {
 7     int m[maxn][maxn];
 8     void init() {
 9         memset(m,0,sizeof m);
10     }
11     void setOne() {
12         init();
13         for(int i = 0; i < maxn; ++i) m[i][i] = 1;
14     }
15     Matrix() {
16         init();
17     }
18     Matrix operator*(const Matrix &t)const {
19         Matrix ret;
20         for(int k = 0; k <= n; ++k) {
21             for(int i = 0; i <= n; ++i)
22                 for(int j = 0; j <= n; ++j)
23                     ret.m[i][j] = (ret.m[i][j] + m[i][k]*t.m[k][j])%mod;
24         }
25         return ret;
26     }
27 };
28 Matrix a,b,c;
29 int main() {
30     int kase;
31     scanf("%d",&kase);
32     while(kase--) {
33         scanf("%d%d",&n,&m);
34         b.init();
35         a.init();
36         c.setOne();
37         for(int i = 1,t,k; i <= n; ++i) {
38             scanf("%d",&t);
39             while(t--) {
40                 scanf("%d",&k);
41                 b.m[k][i] = 1;
42             }
43         }
44         for(int i = 0; i <= n; ++i) {
45             a.m[0][i] = 1;
46             b.m[i][0] = 1;
47         }
48         a.m[0][0] = 0;
49         while(m) {
50             if(m&1) c = c*b;
51             m >>= 1;
52             b = b*b;
53         }
54         a = a*c;
55         printf("%d\n",(a.m[0][0]+1)%mod);
56     }
57     return 0;
58 }