poj3735 Training little cats 矩阵快速幂 强大的应用

#include<iostream>
#include<cstdio>
#include<list>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<cmath>
#include<memory.h>
#include<set>

#define ll long long
#define LL __int64
#define eps 1e-8

//const ll INF=9999999999999;

#define inf 0xfffffff

using namespace std;

//vector<pair<int,int> > G;
//typedef pair<int,int> P;
//vector<pair<int,int>> ::iterator iter;
//
//map<ll,int>mp;
//map<ll,int>::iterator p;
//
//vector<int>G[30012];

typedef struct Node
{
LL m[105][105];
LL r,c;
Node()
{
memset(m,0,sizeof(m));
}
}Matrix;

Matrix multi(Matrix a,Matrix b)//矩阵乘法
{
Matrix c;
c.r=a.r;
c.c=b.c;
for(int i=0;i<a.r;i++)
for(int k=0;k<a.c;k++)
if(a.m[i][k])//这里有个优化，叫做稀疏矩阵优化，百度看定义吧
{
for(int j=0;j<b.c;j++)
c.m[i][j]+=a.m[i][k]*b.m[k][j];
}
return c;
}

int main(void)
{
LL n,m,k;
while(scanf("%I64d %I64d %I64d",&n,&m,&k),n+m+k)
{
Matrix flg,a;
a.c=n+1;
a.r=n+1;
for(int i=0;i<=n;i++)
a.m[i][i]=1;//获得单位矩阵亦是操作矩阵
flg.c=n+1,flg.r=1;
flg.m[0]
=1;//获得花生米矩阵
char s[2];
LL x,y;
while(k--)
{
scanf("%s",s);
if(s[0]=='g')
{
scanf("%I64d",&x);
a.m
[x-1]++;//给花生x列最后一行加
}
if(s[0]=='e')
{
scanf("%I64d",&x);
for(int i=0;i<=n;i++)
a.m[i][x-1]=0;//吃花生x列全清空
}
if(s[0]=='s')
{
scanf("%I64d %I64d",&x,&y);
for(int i=0;i<=n;i++)
{
LL tmp=a.m[i][x-1];
a.m[i][x-1]=a.m[i][y-1];
a.m[i][y-1]=tmp;//x列与y交换
}
}
}
while(m)//相当于快速幂过程，
{
if(m&1)
{
flg=multi(flg,a);
m--;
}
else
{
m>>=1;
a=multi(a,a);
}
}
printf("%I64d",flg.m[0][0]);
for(int i=1;i<n;i++)
printf(" %I64d",flg.m[0][i]);
puts("");
}
}