Training little cats

Training little cats

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总时间限制: 

2000ms 

内存限制: 

65536kB

描述

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.

输入

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)

输出

For each test case, output n numbers in a single line, representing the numbers of peanuts the cats have.

样例输入

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

样例输出

2 0 1

来源

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Facer

将操作用矩阵表示之后二分

#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<iomanip>#include<queue>#include<stack>#include<vector>#include<set>#include<map>using namespace std;long long int Transform[105][105];long long int tmp[105][105];long long int peanut[105]={0};long long int p[105]={0};char c;int main(){int n,m,k,x,y;while(cin>>n>>m>>k){if((n|m|k)==0)break;memset(Transform,0,sizeof(Transform));for(int i=0;i<=n;++i)Transform[i][i]=1;for(int i=0;i<k;++i){cin>>c;if(c=='g'){cin>>x;Transform[x-1][n]++;}else if(c=='e'){cin>>x;for(int i=0;i<=n;++i)Transform[x-1][i]=0;}else if(c=='s'){cin>>x>>y;for(int i=0;i<=n;++i)swap(Transform[x-1][i],Transform[y-1][i]);}}memset(peanut,0,sizeof(peanut));peanut[n]=1;while(m>0){if(m%2==1){memset(p,0,sizeof(p));for(int i=0;i<=n;++i){for(int j=0;j<=n;++j){p[i]+=Transform[i][j]*peanut[j];}}memcpy(peanut,p,(n+1)*sizeof(long long int));}memset(tmp,0,sizeof(tmp));for(int i=0;i<=n;++i){for(int j=0;j<=n;++j){if(Transform[i][j]==0)continue;for(int k=0;k<=n;++k){tmp[i][k]+=Transform[i][j]*Transform[j][k];}}}memcpy(Transform,tmp,sizeof(tmp));m>>=1;}for(int i=0;i<n;++i){cout<<peanut[i]<<" ";}cout<<endl;}return 0;}