HDU2371 矩阵计算转置

转http://blog.csdn.net/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 x₁,x₂,...,x_n be the sequence of characters of the string to be encoded. 

1. Choose an integer m and n pairwise distinct numbers p₁,p₂,...,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 p₁, ..., 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 ≤ 10⁹). The following line consists of n pairwise different numbers p₁,...,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 32 3 1 5 4helol16 80428938413 10 2 7 8 1 16 12 15 6 5 14 3 4 11 9scssoet tcaede n8 125 3 4 2 1 8 6 7encoded?0 0

 

Sample Output

hellosecond test caseencoded?

 

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 falsetypedef 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 100const 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;}