poj 3735Training little cats（矩阵快速幂完成模拟注意优化）

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最新推荐文章于 2024-10-30 23:49:49 发布

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分类专栏：编程算法文章标签：矩阵快速幂 c++ poj

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本文介绍了一种通过矩阵运算来模拟猫咪进行特定训练的算法。该算法能够处理三种基本操作，并通过高效的矩阵幂运算来求解大规模重复操作下每只猫咪拥有的花生数量。

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Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12392		Accepted: 3049

Description

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), Facer

/*
难点在与怎么将这三种操作转换为矩阵的操作，以题目上的测试数据为例，定义表示猫咪的花生的向量A：

1 0 0 0
A[id]表示第id只猫的初始花生数量，初始时猫咪没有花生，所以都是0。当然，首个元素固定是1，这是为了做乘法用的。

定义单位初始方阵T：

1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
定义 R = A * T，那么R[id]表示第id只猫的最终花生数量。上面向量A那个突兀的1就是为了取矩阵T第1行各个值的。

那么

g i   := ++T[0][i]

e i   := set T[i][i] = 0 即将矩阵的元素(i,i)置为0

s i j := swap(T[t][i], T[t][j]) (0<=t<=height) 即将矩阵的i列和j列互换。

T执行m次，结果R = A * Tm 。轮到矩阵的幂出场了，将上次的模板改改即可：
*/
 
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
typedef long long ll;
using namespace std;
struct matrix{
	ll mat[150][150];
	matrix()
	{
		memset(mat,0,sizeof(mat));
	}
};
matrix ret,d;
ll n,m,k;
matrix mul(matrix a,matrix b)
{
	memset(ret.mat,0,sizeof(ret.mat));
	for(int i=0;i<=n;i++)
	{
	
		for(int k=0;k<=n;k++)//优化 防止超时 
		if(a.mat[i][k])
		{
		for(int j=0;j<=n;j++)
		{
			ret.mat[i][j]+=a.mat[i][k]*b.mat[k][j];	
		}	
		}
	}
	return ret;
}
matrix ks(matrix a,long long k)
{
	
	if(k==1)
	return a;
	matrix e;
	for(int i=0;i<=n;i++)
	{
		e.mat[i][i]=1;
	}
	if(k==0)
	return e;
	while(k)
	{
		if(k&1)
		{
			e=mul(e,a);
		}
		a=mul(a,a);
		k>>=1;
	}
	return e;
}
int main()
{
	while(~scanf("%lld%lld%lld",&n,&m,&k))
	{
		if(!n&&!m&&!k)break;
		ll a,b;
		memset(d.mat,0,sizeof(d.mat));
		for(int i=0;i<=n;i++)
		d.mat[i][i]=1;
		char op[5];
		while(k--)
		{
			scanf("%s",op);
			if(op[0]=='g')//  第一行表示每只猫的豆豆数  当输入g的时候a猫的豆豆加一  当输入e时 把a猫的所有豆豆清空  交换时把两列的数据全部交换 
			{
				scanf("%lld",&a);
				d.mat[0][a]++;
			}
			else if(op[0]=='e')
			{
				scanf("%lld",&a);
				for(int i=0;i<=n;i++)
				d.mat[i][a]=0;
			}
			else {
				scanf("%lld%lld",&a,&b);
				for(int i=0;i<=n;i++)
				{
					ll t=d.mat[i][a];
					d.mat[i][a]=d.mat[i][b];
					d.mat[i][b]=t;
				}
			}
		}
		matrix ans=ks(d,m);
		printf("%lld",ans.mat[0][1]);
		for(int i=2;i<=n;i++)
		{
			printf(" %lld",ans.mat[0][i]);
		}
		printf("\n");
	}
	return 0;
}