poj 3735 Training little cats

猫咪训练算法

最新推荐文章于 2018-08-29 16:34:51 发布

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#矩阵快速幂 poj3735

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本文介绍了一种使用矩阵快速幂和矩阵乘法优化的算法，解决猫咪训练中花生分配问题。通过命令序列的重复执行，确定每只猫最终获得的花生数量。

Facer's pet cat just gave birth to a brood of little cats. Having considered the health of those lovely cats, Facer decides to make the cats to do some exercises. Facer has well designed a set of moves for his cats. He is now asking you to supervise the cats to do his exercises. Facer's great exercise for cats contains three different moves:g i : Let the ith cat take a peanut.e i : Let the ith cat eat all peanuts it have.s i j : Let the ith cat and jth cat exchange their peanuts.All the cats perform a sequence of these moves and must repeat it m times! Poor cats! Only Facer can come up with such embarrassing idea. You have to determine the final number of peanuts each cat have, and directly give them the exact quantity in order to save them.

Input

The input file consists of multiple test cases, ending with three zeroes "0 0 0". For each test case, three integers n, m and k are given firstly, where n is the number of cats and k is the length of the move sequence. The following k lines describe the sequence.(m≤1,000,000,000, n≤100, k≤100)

Output

For each test case, output n numbers in a single line, representing the numbers of peanuts the cats have.

Sample Input

Sample Output

2 0 1

Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17)

题目大意：有 n只猫，三种关于花生的命令 (得花生，吃花生，交换花生)，给出一套命令，重复n次，问最后每只喵呜得到多少花生。

思路：矩阵快速幂构造加矩阵乘法优化，网上题解很多自己去看...

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=105;
typedef long long LL;
int N; 
struct Matrix
{
	LL matrix[maxn][maxn];
	Matrix()
	{
		memset(matrix,0,sizeof(matrix)); 
	}
	void init()
	{
		for(int i=0;i<maxn;i++)matrix[i][i]=1;
	}
};
Matrix matrixmul(Matrix a,Matrix b)
{
	Matrix ans;
	for(int i=0;i<N;i++)
	{
		for(int k=0;k<N;k++)
		{
			if(!a.matrix[i][k])continue;
			for(int j=0;j<N;j++)
			{
				ans.matrix[i][j]+=a.matrix[i][k]*b.matrix[k][j];
			}
		}
	}
	return ans;
} 
Matrix matrixquickpow(Matrix a,int b)
{
	Matrix ans;
	ans.init();
	while(b)
	{
		if(b&1)ans=matrixmul(ans,a);
		b>>=1;
		a=matrixmul(a,a);
	}
	return ans;
}
int main()
{
	int n,m,k,x,y;
	char s;
	while(scanf("%d%d%d",&n,&k,&m)==3&&(m||n||k)) 
	{
		Matrix mtr;
		mtr.init();
		N=n+1;
		int cas=1;
		while(m--)
		{
			cin>>s;
			if(s=='g')
			{
				scanf("%d",&x);
				mtr.matrix[x-1][n]++;
			}
			else if(s=='e')
			{
				scanf("%d",&x);
				for(int i=0;i<=n;i++)mtr.matrix[x-1][i]=0;
			}
			else if(s=='s')
			{
				scanf("%d%d",&x,&y);
				for(int i=0;i<=n;i++)swap(mtr.matrix[x-1][i],mtr.matrix[y-1][i]);
			}
		}
		Matrix ans=matrixquickpow(mtr,k);
		for(int i=0;i<n;i++)
		{
			if(i==0)printf("%I64d",ans.matrix[i][n]);
			else printf(" %I64d",ans.matrix[i][n]);
		}
		printf("\n");
	}
}