POJ 3735（矩阵快速幂||稀疏矩阵优化）

https://vjudge.net/contest/145472#problem/C

题意：有n只猫，有三种操作得花生，吃花生，换花生。k种操作，进行m轮

思路：m很大，考虑矩阵变化，考虑每一个变化过程，由于有加一，将初始矩阵末尾增加一，方便进行操作，然后有如下变换

这道题要注意矩阵为稀疏矩阵，应该先判断a矩阵改值是否存在再加入到ans中，否则会TLE

#include<stdio.h>
#include<iostream>
#include<cmath>
#include<string.h>
#define fi first
#define se second
#define log2(a) log(n)/log(2)
#define show(a) cout<<a<<endl;
#define show2(a,b) cout<<a<<" "<<b<<endl;
#define show3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl;
using namespace std;

typedef long long ll;

const ll inf = 1e17 + 10;
const int N = 1e6 + 10;
const ll mod = 1e9 + 7;
const int base = 131;
const double pi = acos ( -1 );




ll n, m, k;
struct mat
{
	ll x[105][105];
	mat()
	{
		memset ( x, 0, sizeof x );
	}
};
mat ans, tr, a;

mat mul ( mat a, mat b )
{
	mat ans;
	for ( int i = 1; i <= n + 1; i++ )
	{
		for ( int k = 1; k <= n + 1; k++ )
		{
			if ( a.x[i][k] )
				for ( int j = 1; j <= n + 1; j++ )
				{
					ans.x[i][j] = ( ans.x[i][j] + a.x[i][k] * b.x[k][j] );
				}
		}
	}
	return ans;
}

mat ksm ( mat x, ll k )
{
	mat ans;
	for ( int i = 1; i <= n + 1; i++ ) ans.x[i][i] = 1;
	while ( k )
	{
		if ( k & 1 ) ans = mul ( ans, x );
		k >>= 1;
		x = mul ( x, x );
	}
	return ans;
}


int main( )
{
	while ( scanf ( "%lld %lld %lld", &n, &m, &k ) )
	{
		if ( !n && !m && !k ) break;
		for ( int i = 1; i <= n + 1; i++ )
		{
			ans.x[i][1] = 0;
			for ( int j = 1; j <= n + 1; j++ )
			{
				tr.x[i][j] = ( i == j ) ? 1 : 0;
			}
		}
		ans.x[n + 1][1] = 1;

		for ( int i = 1; i <= k; i++ )
		{
			//show2("k",i)
			char ch;
			int x, y;
			getchar();
			scanf ( "%c%d", &ch, &x );

			if ( ch == 'g' )
			{


				tr.x[x][n + 1]++;
			}
			else if ( ch == 'e' )
			{

				for ( int i = 1; i <= n + 1; i++ ) tr.x[x][i] = 0;
			}
			else
			{
				scanf ( "%d", &y );
				for ( int i = 1; i <= n + 1; i++ ) swap ( tr.x[x][i], tr.x[y][i] );
			}
		}


		ans = mul ( ksm ( tr, m ), ans );
		for ( int i = 1; i < n; i++ )
		{
			printf ( "%lld ", ans.x[i][1] );
		}
		printf ( "%lld\n", ans.x[n][1] );
	}

}