poj 3735 Training little cats

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 6g 1g 2g 2s 1 2g 3e 20 0 0

Sample Output

2 0 1

Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17)

题目大意：有n只猫，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。

思路：矩阵快速幂构造加矩阵乘法优化，网上题解很多自己去看...

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn=105;typedef long long LL;int N; struct Matrix{LL matrix[maxn][maxn];Matrix(){memset(matrix,0,sizeof(matrix)); }void init(){for(int i=0;i<maxn;i++)matrix[i][i]=1;}};Matrix matrixmul(Matrix a,Matrix b){Matrix ans;for(int i=0;i<N;i++){for(int k=0;k<N;k++){if(!a.matrix[i][k])continue;for(int j=0;j<N;j++){ans.matrix[i][j]+=a.matrix[i][k]*b.matrix[k][j];}}}return ans;} Matrix matrixquickpow(Matrix a,int b){Matrix ans;ans.init();while(b){if(b&1)ans=matrixmul(ans,a);b>>=1;a=matrixmul(a,a);}return ans;}int main(){int n,m,k,x,y;char s;while(scanf("%d%d%d",&n,&k,&m)==3&&(m||n||k)) {Matrix mtr;mtr.init();N=n+1;int cas=1;while(m--){cin>>s;if(s=='g'){scanf("%d",&x);mtr.matrix[x-1][n]++;}else if(s=='e'){scanf("%d",&x);for(int i=0;i<=n;i++)mtr.matrix[x-1][i]=0;}else if(s=='s'){scanf("%d%d",&x,&y);for(int i=0;i<=n;i++)swap(mtr.matrix[x-1][i],mtr.matrix[y-1][i]);}}Matrix ans=matrixquickpow(mtr,k);for(int i=0;i<n;i++){if(i==0)printf("%I64d",ans.matrix[i][n]);else printf(" %I64d",ans.matrix[i][n]);}printf("\n");}}