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POJ 3735 Training little cats 解题报告（矩阵构造+快速幂优化）

2014-04-24 21:47 525 查看
Training little cats

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 9396 Accepted: 2249
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
3 1 6
g 1
g 2
g 2
s 1 2
g 3
e 2
0 0 0


Sample Output
2 0 1


Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Facer
   

    解题报告： WA/TLE很久的一道题。

    用long long TLE， unsigned WA，纠结至死……无奈看Discuss，发现了惊人的答案：矩阵乘法优化。其实以前就知道的。如果某一行很多个0，可以先判断是否为0，是就不用乘了。这样乘法的复杂度从100W直接降到1W左右（每行最多两个数不为0），乘上m的快速幂也不会超时了。

    代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int SIZE=101;
typedef long long LL;
int n;

struct Matrix
{
LL a[SIZE][SIZE];

Matrix(int t=0)
{
memset(a, 0, sizeof(a));
for(int i=0;i<SIZE;i++) a[i][i]=t;
}

Matrix operator*(const Matrix& b) const
{
Matrix c;
for(int k=0;k<=n;k++)
for(int i=0;i<=n;i++) if(a[i][k])
for(int j=0;j<=n;j++) if(b.a[k][j])
c.a[i][j]+=a[i][k]*b.a[k][j];
return c;
}
};

Matrix powM(Matrix a, int b)
{
Matrix res(1);
while(b)
{
if(b&1)
res=res*a;
a=a*a;
b>>=1;
}
return res;
}

int main()
{
int m, k;
while(~scanf("%d%d%d", &n, &m, &k) && (n||m||k))
{
Matrix op(1);
while(k--)
{
char ch[2];
scanf("%s", ch);

if(ch[0]=='g')
{
int t;
scanf("%d", &t);
op.a[0][t]++;
}
else if(ch[0]=='e')
{
int t;
scanf("%d", &t);
for(int i=0;i<=n;i++)
op.a[i][t]=0;
}
else
{
int a, b;
scanf("%d%d", &a, &b);
for(int i=0;i<=n;i++)
swap(op.a[i][a], op.a[i][b]);
}
}

op=powM(op, m);

for(int i=1;i<=n;i++)
printf("%lld ", op.a[0][i]);
puts("");
}
}
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标签：  矩阵乘法优化
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