Training little cats(矩阵快速幂)

简介

题目来源:POJ3735

Facer’s pet cat just gave birth to a brood of little cats. Having considered the health of those lovely cats, Facer decides to make the cats to do some exercises. Facer has well designed a set of moves for his cats. He is now asking you to supervise the cats to do his exercises. Facer’s great exercise for cats contains three different moves:

  • g i : Let the ith cat take a peanut.
  • e i : Let the ith cat eat all peanuts it have.
  • s i j : Let the ith cat and jth cat exchange their peanuts.

All the cats perform a sequence of these moves and must repeat it m times! Poor cats! Only Facer can come up with such embarrassing idea.
You have to determine the final number of peanuts each cat have, and directly give them the exact quantity in order to save them.

Input

The input file consists of multiple test cases, ending with three zeroes “$0 0 0$”. For each test case, three integers $n$, $m$ and $k$ are given firstly, where $n$ is the number of cats and $k$ is the length of the move sequence. The following $k$ lines describe the sequence.
$(m≤1,000,000,000, n≤100, k≤100)$

Output

For each test case, output $n$ numbers in a single line, representing the numbers of peanuts the cats have.

Sample Input

1
2
3
4
5
6
7
8
3 1 6
g 1
g 2
g 2
s 1 2
g 3
e 2
0 0 0

Sample Output

1
2 0 1

思路

  • 每种操作对应矩阵的一种变换。以$3$只猫为例:
    • 初始列向量为$(0,0,0,1)^T$,转移矩阵为单位阵。
  • 每种操作对应矩阵的一个变换。
    • g操作即该行最后一列元素加$1$;
    • e操作即改行元素全部清零;
    • s操作即换法变换。
  • 构造完矩阵之后矩阵快速幂即可。
  • 注意稀疏矩阵在矩阵乘法中的优化。

代码

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#include <cstdio>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector <ll> vec;
typedef vector <vec> mat;

mat mul(mat &A, mat &B)
{
mat C(A.size(), vec(B[0].size()));
for (int i = 0; i < A.size(); i++)
for (int k = 0; k < B.size(); k++)
if (A[i][k])
for (int j = 0; j < B[0].size(); j++)
C[i][j] += A[i][k] * B[k][j];
return C;
}

mat fast_pow (mat A, ll n)
{
mat B(A.size(), vec(A.size()));
for (int i = 0; i < A.size(); i++)
B[i][i] = 1;
while (n)
{
if (n & 1)
B = mul(B, A);
A = mul(A, A);
n >>= 1;
}
return B;
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("1.in", "r", stdin);
freopen("1.out", "w", stdout);
#endif
int n, m, k;
while (scanf("%d%d%d", &n, &m, &k), n || m || k)
{
mat A(n + 1, vec(n + 1));
for (int i = 0; i < A.size(); i++)
A[i][i] = 1;
while (k--)
{
char cmd[2];
scanf("%s", cmd);
if (cmd[0] == 'g')
{
int id;
scanf("%d", &id);
A[id - 1][n]++;
}
else if (cmd[0] == 'e')
{
int id;
scanf("%d", &id);
for (int i = 0; i <= n; i++)
A[id - 1][i] = 0;
}
else
{
int a, b;
scanf("%d%d", &a, &b);
for (int i = 0; i <= n; i++)
swap(A[a - 1][i], A[b - 1][i]);
}
}
A = fast_pow(A, m);
vec col(n + 1);
vec ans(n + 1);
col[n] = 1;
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= n; j++)
ans[i] += A[i][j] * col[j];
if (i)
putchar(' ');
if (i != n)
printf("%lld", ans[i]);
}
putchar('\n');
}
return 0;
}
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