poj 3735 Training little cats（构造矩阵）

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#define LL long long
#define _LL __int64
#define eps 1e-8
#define PI acos(-1.0)

using namespace std;
const int maxn = 105;

int n,k,m;

struct matrix
{
	_LL mat[maxn][maxn];
	//初始化为单位矩阵
	void init()
	{
		memset(mat,0,sizeof(mat));
		for(int i = 0; i <= 102; i++) //不知道为什么，102换成maxn就会崩，求解。
			mat[i][i] = 1;
	}
}a,res;

//矩阵相乘
matrix matrixMul(matrix x, matrix y)
{
	matrix tmp;
	memset(tmp.mat,0,sizeof(tmp.mat));

	for(int i = 0; i <= n; i++)
	{
		for(int k = 0; k <= n; k++)
		{
			if(x.mat[i][k] == 0) continue;
			for(int j = 0; j <= n; j++)
			{
				tmp.mat[i][j] += x.mat[i][k] * y.mat[k][j];
			}
		}
	}
	return tmp;
}

//矩阵求幂
matrix Mul(matrix x, int k)
{

	matrix tmp;
	tmp.init();
	
	while(k)
	{
		if(k & 1)
			tmp = matrixMul(tmp,x);
		x = matrixMul(x,x);
		k >>= 1;
	}
	return tmp;
}

int main()
{
	char ch[4];

	while(~scanf("%d %d %d",&n,&m,&k))
	{
		if(n == 0 && k == 0 && m == 0) break;
		a.init();

        int num1,num2;
        while(k--)
        {
            scanf("%s",ch);
            if(ch[0] == 'g')
            {
                scanf("%d",&num1);
                a.mat[0][num1] += 1;
            }
            else if(ch[0] == 'e')
            {
                scanf("%d",&num1);
                for(int i = 0; i <= n; i++)
                    a.mat[i][num1] = 0;
            }
            else
            {
                scanf("%d %d",&num1,&num2);
                for(int i = 0; i <= n; i++)
                    swap(a.mat[i][num1],a.mat[i][num2]);
            }
        }
		if(m == 0)
		{
			for(int i = 0; i < n;i++)
			{
				if(i == 0)printf("0");
				else printf(" 0");
			}
            printf("\n");
            continue ;
		}
        res = Mul(a,m);
        for(int i = 1; i < n; i++)
			printf("%I64d ",res.mat[0][i]);
		printf("%I64d\n",res.mat[0][n]);
	}

	return 0;
}