HDU 5411 CRB and Puzzle
来源:互联网 发布:程序员的硬技能 编辑:程序博客网 时间:2024/05/29 19:48
Problem Description
CRB is now playing Jigsaw Puzzle.
There areN 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 mostM 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 .)
There are
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
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 integersN , M denoting the number of kinds of pieces and the maximum number of moves.
ThenN lines follow. i -th line is described as following format.
ka1 a2 ... ak
Herek 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 ≤T ≤ 20
1 ≤N ≤ 50
1 ≤M ≤ 105
0 ≤k ≤ N
1 ≤a1 < a2 < … < ak ≤ N
The first line contains two integers
Then
k
Here
1 ≤
1 ≤
1 ≤
0 ≤
1 ≤
Output
For each test case, output a single integer - number of different patterns modulo 2015.
Sample Input
13 21 21 30
Sample Output
6Hintpossible patterns are ∅, 1, 2, 3, 1→2, 2→3找到递推式,来一发矩阵乘法就ok了
#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<cstdio>#include<ctime>#include<vector>using namespace std;const int base = 2015;const int maxn = 100005;int T, n, m, k, kk;struct Matrix{ #define size 55 int m[size][size]; Matrix() { memset(m, 0, sizeof(m)); } void operator =(const Matrix&b) { memcpy(m, b.m, sizeof(m)); } Matrix operator *(const Matrix&b) { Matrix c; for (int i = 0; i < size; i++) for (int j = 0; j < size; j++) for (int k = 0; k < size; k++) c.m[i][k] = (c.m[i][k] + m[i][j] * b.m[j][k]) % base; return c; } Matrix get(int x) { Matrix a, b = *this; for (int f = 0; x; x >>= 1) { if (x & 1) if (f) a = a * b; else a = b, f = 1; b = b * b; } return a; }};int main(){ scanf("%d", &T); while (T--) { scanf("%d%d", &n, &m); Matrix a, b; for (int i = 1; i <= n + 1; i++) a.m[i][n + 1] = 1; for (int i = 1; i <= n; i++) { scanf("%d", &k); for (int j = 1; j <= k; j++) { scanf("%d", &kk); a.m[i][kk] = 1; } } for (int i = 1; i <= n + 1; i++) b.m[1][i] = 1; a = a.get(m); b = b * a; if (m == 1) printf("%d\n", n + 1); else printf("%d\n", b.m[1][n + 1]); } return 0;}
0 0
- HDU 5411 CRB and Puzzle
- hdu 5411 CRB and Puzzle
- hdu 5411 CRB and Puzzle 矩阵快速幂
- 【矩阵快速幂】 HDU 5411 CRB and Puzzle 等比
- hdu 5411 CRB and Puzzle (矩阵快速幂优化dp)
- hdu 5411 CRB and Puzzle(矩阵快速幂)
- hdu 5411 CRB and Puzzle(矩阵快速幂)
- HDU 5411 CRB and puzzle (Dp + 矩阵快速幂)
- hdu 5411 CRB and Puzzle(矩阵快速幂)
- hdu 5411 CRB and Puzzle 矩阵快速幂
- hdu 5411 CRB and Puzzle【矩阵快速幂】
- 矩阵快速幂(CRB and Puzzle,HDU 5411)
- HHDU5411 CRB and Puzzle
- hdu 5411 CRB and Puzzle 2015 多校联合训练赛#10 快速矩阵幂
- HDU 5411 CRB and Puzzle (2015年多校比赛第10场)
- HDU 5411 CRB and Puzzle(矩阵快速幂+可达矩阵)
- HDU 5411(CRB and Puzzle-矩阵A+A^2+..+A^n)
- [HDU 5411] CRB and Puzzle (矩阵加速DP + 前缀和矩阵|等比求和快速幂 )
- Android开发最佳实践
- DOM(二)-12-(示例-表格排序)
- Spring AOP框架学习笔记(2):AOP拦截器调用的实现
- 迟来的Robocup2015总结及15国赛简单预测
- SQLServer索引的四个高级特性
- HDU 5411 CRB and Puzzle
- Oracle自增序列
- 2015年8月20日17:15:53
- Android中AsyncTask的简单用法
- Android中几种关闭Activity或app的方法
- 收藏下 Unicode范围
- Linux命令more
- SAP调用webservice 参数为空
- UVA 529 Addition Chains