Training little cats

5 Training little cats

题目

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

思路

m太大了，要想重复这么多次，只能构建矩阵，用矩阵快速幂

让n++，构建n*n的 ‘’1‘’ 矩阵，最后一列的前n-1存答案，g就在a [b] [n] ++; e,就让该行都为0，s，就swap交换这两行

而且较为稀疏，要再进行优化

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    long long cp[110][110];
} s;
int n;
node juzhen(node a,node b)
{
    node c;
    memset(c.cp,0,sizeof(c.cp));
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(a.cp[i][j])//优化必不可少，否则会超时
            {
                for(int z=1; z<=n; z++)
                    c.cp[i][z]+=a.cp[i][j]*b.cp[j][z];
            }
        }
    }
    return c;
}
node mul(int m)
{
    node ans;
    memset(ans.cp,0,sizeof(ans.cp));
    for(int i=1; i<=n; i++)
        ans.cp[i][i]=1;
    while(m)
    {
        if(m%2)
            ans=juzhen(ans,s);
        m>>=1;
        s=juzhen(s,s);
    }
    return ans;
}
int main()
{
    int k,b,c;
    long long m;
    char a[10];
    while(~scanf("%d%lld%d",&n,&m,&k))
    {
        if(n==m&&m==k&&k==0)
            break;
        n++;
        memset(s.cp,0,sizeof(s.cp));
        for(int i=1; i<=n; i++)
            s.cp[i][i]=1;
        while(k--)
        {
            scanf("%s%d",a,&b);
            if(a[0]=='g')
            {
                s.cp[b][n]++;
            }
            else if(a[0]=='e')
            {
                for(int j=1; j<=n; j++)
                    s.cp[b][j]=0;
            }
            else
            {
                scanf("%d",&c);
                for(int j=1; j<=n; j++)
                {
                    long long d;
                    d=s.cp[b][j];
                    s.cp[b][j]=s.cp[c][j];
                    s.cp[c][j]=d;
                }
            }
        }
        node p=mul(m);
        for(int i=1; i<n; i++)
            printf("%lld ",p.cp[i][n]);
        printf("\n");
    }
    return 0;
}