poj3735-Training little cats

poj3735 Training little cats

Training little cats
Time Limit: 2000MS

Memory Limit: 65536K
Total Submissions: 14750

Accepted: 3648
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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 105;
typedef long long LL;

struct Matrix{
    LL n, m;
    LL a[MAXN][MAXN];
    void clear(){
      n = m = 0;
      memset(a, 0, sizeof(a));
    }
    Matrix operator * (const Matrix &b) const{
      Matrix tmp;
      tmp.clear();
      for(int i = 1; i <= m; i++)
        for(int k = 1; k <= m; k++)
          if(a[i][k])
            for(int j = 1; j <= m; j++)
              tmp.a[i][j] += a[i][k] * b.a[k][j];   
      tmp.n = n, tmp.m = m;
      return tmp;
    }
}S, A, I;

Matrix Fast(LL x){
    Matrix ans = I;
    while(x){
      if(x & 1) ans = ans * A;
      A = A * A;
      x >>= 1; 
    }
    return ans;
} 

void init(int n){
    A.clear(); S.clear(); I.clear();
    A.n = n+1, A.m = n+1;
    S.n = 1, S.m = n+1,S.a[1][n+1] = 1;
    I.n = n+1, I.m = n+1; 
    for(int i = 1; i<= n+1; i++) A.a[i][i] = 1;
    for(int i = 1; i <= n+1; i++) I.a[i][i] = 1;
}

int main()
{
    LL n, m, k;
    while((~scanf("%lld%lld%lld", &n, &m, &k)) && n){
      init(n);
      LL qi, qj;
      char qus[5];
      for(int i = 1; i <= k; i++){
        scanf("%s", qus);
        if(qus[0] == 'g'){
          scanf("%lld", &qi);
          A.a[n+1][qi]++;
        }
        else if(qus[0] == 'e'){
          scanf("%lld", &qi);
          for(int i = 1; i <= n+1; i++) A.a[i][qi] = 0;
        }
        else{
          scanf("%lld%lld", &qi, &qj);
          for(int i = 1; i <= n+1; i++)
            swap(A.a[i][qi], A.a[i][qj]);
        }
      }
      Matrix ans = S * Fast(m);
      for(int i = 1; i <= n; i++)
        printf("%lld ", ans.a[1][i]);
      printf("\n");
    }
    return 0;
}

注意到m的范围很大而n的最大取值只有100，所以想到主要的思路：快速幂 S * （A ^ m）
为了能支持添加操作，故作一个1*(n+1)的矩阵，初始时（1，1）~（1，n） = 0，（1， n+1）=1用于进行加法。
三个操作中比较难想的是删除操作，其实只需要把矩阵中对应列的所有项清空即可。后续添加操作在第n+1行进行，而做矩阵乘法时，该列与S相乘便不再添加上一步的糖果数。
仅仅这样会tle
注意到矩阵A有很多的0，要用稀松矩阵乘法，即：

Matrix operator * (const Matrix &b) const{
      Matrix tmp;
      tmp.clear();
      for(int i = 1; i <= m; i++)
        for(int k = 1; k <= m; k++)
          if(a[i][k])
            for(int j = 1; j <= m; j++)
              tmp.a[i][j] += a[i][k] * b.a[k][j];   
      tmp.n = n, tmp.m = m;
      return tmp;
    }n tmp;
    }

即对Matrix tmp, a, b
tmp = a * b 只有a[i][j] != 0时把所有tmp中与a[i][j]有关的相加上a与b对应的乘积。