HDU2371 矩阵计算转置

转http://blog.youkuaiyun.com/xingyeyongheng/article/details/9855103

Decode the Strings

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 766    Accepted Submission(s): 232

Problem Description

Bruce Force has had an interesting idea how to encode strings. The following is the description of how the encoding is done: 

Let x ₁,x ₂,...,x _n be the sequence of characters of the string to be encoded. 

1. Choose an integer m and n pairwise distinct numbers p ₁,p ₂,...,p _n from the set {1, 2, ..., n} (a permutation of the numbers 1 to n). 
2. Repeat the following step m times. 
3. For 1 ≤ i ≤ n set y _i to x _pi, and then for 1 ≤ i ≤ n replace x _i by y _i. 

For example, when we want to encode the string "hello", and we choose the value m = 3 and the permutation 2, 3, 1, 5, 4, the data would be encoded in 3 steps: "hello" -> "elhol" -> "lhelo" -> "helol". 

Bruce gives you the encoded strings, and the numbers m and p ₁, ..., p _n used to encode these strings. He claims that because he used huge numbers m for encoding, you will need a lot of time to decode the strings. Can you disprove this claim by quickly decoding the strings? 

 

Input

The input contains several test cases. Each test case starts with a line containing two numbers n and m (1 ≤ n ≤ 80, 1 ≤ m ≤ 10 ⁹). The following line consists of n pairwise different numbers p ₁,...,p _n (1 ≤ p _i ≤ n). The third line of each test case consists of exactly n characters, and represent the encoded string. The last test case is followed by a line containing two zeros. 

 

Output

For each test case, print one line with the decoded string. 

 

Sample Input

  

   5 3
2 3 1 5 4
helol
16 804289384
13 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9
scssoet tcaede n
8 12
5 3 4 2 1 8 6 7
encoded?
0 0
  

 

Sample Output

  

   hello
second test case
encoded?
  

 

 

  题目意思是给出n个字符的置换方式,经过m次转换后得到了最终字符串(就是给定的字符串),求最初的字符串
 
 

  分析:假定最初字符串序号是1,2,3,4,置换方式是3,1,2,4，即1,2,3,4置换一次后得到3,1,2,4构成的字符串
 
 

   将1 2 3 4置换为3 1 2 4，相当于下面的矩阵乘法：
  

       
  
 
 

  如果置换m次则将置换矩阵*m次即可，最后乘上给定的字符串矩阵得到最终字符串矩阵(就是得到的矩阵第i行第j列是1表示由s[j]得到第s[i]个字符)
 
 

  但是本题是给定结果，叫我们求最初的矩阵，其实就是将原来矩阵求逆矩阵的m次
 
 

  A*B^m=C =>A=C*B^(-m),
 
 

  A*A^-1=I;//I是单位矩阵
 
 

  注意到这里的矩阵A元素为0或1且每一行每一列只有一个1，则A中的A[i][k]*B[k][j]=I[i][j]=1,i == j,所以A的逆矩阵就是A逆s[i][j]=A的s[j][i]
 

#define DeBUG
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <sstream>
#include <map>
#include <list>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 0x3f3f3f3f
#define EPS 1e-6
#define TRUE true
#define FALSE false
typedef long long LL;
const double PI = acos(-1.0);
//#pragma comment(linker, "/STACK:102400000,102400000")
inline int sgn(double x)
{
    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);
}
#define N 100
const int MAXN = 100;
long long mod;
struct Matrix
{
    int mat[MAXN][MAXN];
    void Zero()
    {
        memset(mat, 0, sizeof(mat));
    }
    void Unit()
    {
        memset(mat, 0, sizeof(mat));
        for (int i = 0; i < MAXN; i++)
            mat[i][i] = 1;
    }
    void output()
    {
        int from = 1, to = 10;
        for (int i = from - 1; i < to; i++)
        {
            printf("%d ", i);
        }
        printf("\n");
        for (int i = from; i < to; i++)
        {
            printf("%d=", i);
            for (int j = from; j < to; j++)
            {
                printf("%d ", mat[i][j]);
            }
            printf("\n");
        }
    }
};

Matrix operator*(Matrix &a, Matrix &b)
{
    Matrix tmp;
    tmp.Zero();
    for (int k = 0; k < MAXN; k++)
    {
        for (int i = 0; i < MAXN; i++)
        {
            if (!a.mat[i][k])
                continue;
            for (int j = 0; j < MAXN; j++)
            {
                tmp.mat[i][j] += a.mat[i][k] * b.mat[k][j] % mod;
                if ( tmp.mat[i][j] >= mod)
                    tmp.mat[i][j] -= mod;
            }

        }
    }
    return tmp;
}
Matrix operator ^(Matrix a, int k)
{
    Matrix tmp;
    tmp.Unit();
    for (; k; k >>= 1)
    {
        if (k & 1)
            tmp = tmp * a;
        a = a * a;
    }
    return tmp;
}

int a[N];
char s[1000];
Matrix mt;
int main()
{
#ifdef DeBUGs
    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);
#endif
    int n, m;
    mod = 2;
    while (scanf("%d%d", &n, &m), n || m)
    {
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
        }
        getchar();
        gets(s + 1);
        mt.Zero();
        for (int i = 1; i <= n; i++)
        {
            mt.mat[a[i]][i] = 1;
        }
        mt = mt ^ m;
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (mt.mat[i][j])
                {
                    printf("%c", s[j]);
                    break;
                }
            }
        }
        printf("\n");
        // mt.output();
    }
    return 0;
}