POJ 3735 Training little cats<矩阵快速幂/稀疏矩阵的优化>

Training little cats

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13488		Accepted: 3335

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

 

/*
题意：多组输入n,m,k。(m≤1,000,000,000, n≤100, k≤100)，n表示猫的数量，m代表重复的次数，k表示k次操作。
操作种类：
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.
解题过程：
   原本以为是一道简单的模拟题，但是m非常的大，会TLE。看了题解，用的是快速幂，第一次写快速幂，做下总结。
 http://www.hankcs.com/program/algorithm/poj-3735-training-little-cats-time.html
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long  LL;
const int maxn=100+5;
struct node
{
    LL a[maxn][maxn];
};
LL n,m,k;
char s[5];
LL x,y;
node multiplay(node a,node b)
{
    node c;
    memset(c.a, 0, sizeof(c.a));
    for(int i=1;i<=n+1;i++)
        for(int j=1;j<=n+1;j++)
        {
            if(a.a[i][j])
            {
                for(int k=1;k<=n+1;k++)
                    c.a[i][k]+=a.a[i][j]*b.a[j][k];
            }
        }
    return c;
}
node M(node a,LL x)
{
    node c;
    memset(c.a, 0, sizeof(c.a));
    for(int i=1;i<=n+1;i++)
        c.a[i][i]=1;
    while(x)
    {
        if(x&1)
            c=multiplay(c, a);
        a=multiplay(a, a);
        x>>=1;
    }
    return c;
}
int main ()
{
    while(~scanf("%lld%lld%lld",&n,&m,&k))
    {
        if(n==0&&m==0&&k==0)    break;
        node A;
        memset(A.a, 0, sizeof(A.a));
        for(int i=1;i<=n+1;i++)
            A.a[i][i]=1;
        for(int i=1;i<=k;i++)
        {
            scanf("%s",s);
            if(s[0]=='g')
            {
                scanf("%lld",&x);
                A.a[x][n+1]+=1;
            }
            else if(s[0]=='e')
            {
                scanf("%lld",&x);
                memset(A.a[x], 0, sizeof(A.a[x]));
            }
            else
            {
                scanf("%lld%lld",&x,&y);
                for(int i=1;i<=n+1;i++)
                    swap(A.a[x][i], A.a[y][i]);
            }
        }
        A=M(A,m);
        node ans;
        memset(ans.a, 0, sizeof(ans.a));
        ans.a[n+1][1]=1;
        ans=multiplay(A, ans);
        for(int i=1;i<=n;i++)
        {
            if(i==1)
                printf("%lld",ans.a[i][1]);
            else
                printf("% lld",ans.a[i][1]);
        }
        printf("\n");
    }
    return 0;
}