矩阵快速幂 POJ 3735 Training little cats

题目传送门

 /*
     题意：k次操作，g：i猫+1, e：i猫eat，s：swap
     矩阵快速幂：写个转置矩阵，将k次操作写在第0行，定义A = {1，0, 0, 0...}除了第一个外其他是猫的初始值
         自己讲太麻烦了，网上有人讲的很清楚，膜拜之
     详细解释：http://www.cppblog.com/y346491470/articles/157284.html
 */
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 const int MAXN = 1e2 + ;
 const int INF = 0x3f3f3f3f;
 struct Mat  {
     ll m[MAXN][MAXN];
     Mat ()  {
         memset (m, , sizeof (m));
     }
     void init(void) {
         for (int i=; i<MAXN; ++i)  m[i][i] = ;
     }
 };
 int n, m, k;

 Mat operator * (Mat &a, Mat &b) {
     Mat ret;
     for (int k=; k<=n; ++k)    {
         for (int i=; i<=n; ++i)    {
             if (a.m[i][k])  {
                 for (int j=; j<=n; ++j)    {
                     ret.m[i][j] += a.m[i][k] * b.m[k][j];
                 }
             }
         }
     }
     return ret;
 }

 Mat operator ^ (Mat x, int n)   {
     Mat ret;    ret.init ();
     while (n)   {
         if (n & )  ret = ret * x;
         x = x * x;
         n >>= ;
     }
     return ret;
 }

 int main(void)  {       //POJ 3735 Training little cats
     while (scanf ("%d%d%d", &n, &m, &k) == )   {
         if (!n && !m && !k) break;
         Mat A, T;   A.m[][] = ;  T.init ();
         char op[]; int p, q;
         while (k--) {
             scanf ("%s", op);
             if (op[] == 'g')    {
                 scanf ("%d", &p);   T.m[][p]++;
             }
             else if (op[] == 'e')  {
                 scanf ("%d", &p);
                 for (int i=; i<=n; ++i)    T.m[i][p] = ;
             }
             else    {
                 scanf ("%d%d", &p, &q);
                 for (int i=; i<=n; ++i)    swap (T.m[i][p], T.m[i][q]);
             }
         }
         Mat ans = A * (T ^ m);
         for (int i=; i<=n; ++i)    printf ("%I64d%c", ans.m[][i], (i==n) ? '\n' : ' ');
     }

     return ;
 }