poj 3735 Training little cats（构造矩阵）

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#矩阵快速幂

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本文介绍了一种通过矩阵运算来高效计算猫咪训练中花生分配的算法。面对一群小猫进行特定指令训练的问题，文章提出了一种利用矩阵快速幂的方法来解决重复序列指令下每只猫最终拥有的花生数量。

Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8326		Accepted: 2004

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

详细讲解：点击打开链接

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdlib>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define ll long long
using namespace std;

const int maxn=105;

struct Mat
{
    ll mat[maxn][maxn];
    void zero(){
        memset(mat,0,sizeof(mat));
    }
    void unit(){
        zero();
        for(int i=0;i<maxn;i++)
        mat[i][i]=1;
    }
}A,T;
int n,m,k;
Mat operator *(const Mat&a,const Mat &b){
    Mat tmp;
    tmp.zero();
    for(int k=0;k<=n;k++)
    for(int i=0;i<=n;i++)
    if(a.mat[i][k])
    {
        for(int j=0;j<=n;j++)
        tmp.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
    }
    return tmp;
}
Mat operator ^(Mat x,int n){
    Mat tmp;
    tmp.unit();
    while(n)
    {
        if(n&1) tmp=tmp*x;
        x=x*x;
        n>>=1;
    }
    return tmp;
}
void init()
{
    A.zero();
    A.mat[0][0]=1;
    T.unit();
}
int main()
{
    int a,b;
    char s[5];
    while(scanf("%d%d%d",&n,&m,&k),n||m||k)
    {
        init();
        while(k--)
        {
            scanf("%s",s);
            if(s[0]=='g'){
                scanf("%d",&a);
                T.mat[0][a]++;
            }
            else if(s[0]=='e'){
                scanf("%d",&a);
                for(int i=0;i<=n;i++)
                T.mat[i][a]=0;
            }
            else{
                scanf("%d%d",&a,&b);
                for(int i=0;i<=n;i++)
                swap(T.mat[i][a],T.mat[i][b]);
            }
        }
        Mat ans=A*(T^m);
        for(int i=1;i<=n;i++)
        printf("%lld ",ans.mat[0][i]);
        printf("\n");
    }
    return 0;
}