POJ 3745 : Training little cats(矩阵快速幂加优化)

总时间限制:

2000ms

内存限制:

65536kB

描述

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.

输入

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)

输出

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

样例输入

3 1 6
g 1
g 2
g 2
s 1 2
g 3
e 2
0 0 0

样例输出

2 0 1

来源

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

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PKU Campus 2009 (POJ Monthly Contest – 2009.05.17)  

这题题意 如下，有n 只小猫，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。

先构造一个单位矩阵……

然后算矩阵的m 次方，用快速幂， 超时 

还有一个对于稀疏矩阵的加速办法,一般矩阵乘法这么写的：

for (int i=0;i<N;i++)

for (int j=0;j<N;j++)

for (int k=0;k<N;k++)

a[i][j]+=b[i][k]*c[k][j];

改成

for (int i=0;i<N;i++)

for (int j=0;j<N;j++)

if (b[i][j])

for (int k=0;k<N;k++)

a[i][k]+=b[i][j]*c[j][k];

直接过了……另外，注意用 long long

 

 

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

#define maxn 111

long long a[maxn][maxn], b[maxn][maxn],c[maxn][maxn];
int i,j,e,n,m,k;
char op[9];

void Multi(long long a[maxn][maxn],long long b[maxn][maxn])
{
    memset(c,0,sizeof(c));
    int i, j, k;
    for(i=1;i<=n+1;++i)
        for(k=1;k<=n+1;++k)
            if(a[i][k])
                for(j=1;j<=n+1;++j)
                    if(b[k][j])
                        c[i][j]+=a[i][k]*b[k][j];

    for(i=1;i<=n+1;++i)
        for(j=1;j<=n+1;++j)
            a[i][j]=c[i][j];
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k) && !(n==0&&m==0&&k==0))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1;i<=n+1;++i)
            a[i][i]=b[i][i]=1;
        while(k--)
        {
            scanf("%s%d",op,&i);
            if(op[0]=='g')
                ++a[i][n+1];
            if(op[0]=='e')
                for(j=0;j<=n+1;++j)
                    a[i][j]=0;
            if(op[0]=='s')
            {
                scanf("%d",&j);
                for(e=1;e<=n+1;++e)
                    swap(a[i][e],a[j][e]);
            }
        }
        while(m)
        {
            if(m&1)
                Multi(b,a);
            Multi(a,a);
            m>>=1;
        }
        for(i=1;i<=n;++i)
        {
            printf("%lld ",b[i][n+1]);
        }
        printf("\n");
    }
    return 0;
}