POJ 3735 Training little cats

很神奇的一道矩阵题，这也说明了矩阵的神奇之处。  构造矩阵的方法，我是想不出来了，只能从网上找了找思路。

因m的数据范围较大，用矩阵连乘。
构建矩阵模型，peanut[N] = {0,0，。。。。0,1}：即前n个数为0,最后一个数取1
matrix[N][N],初始化条件下为单位矩阵，。。。
对猫咪进行操作转化为在对矩阵peanut进行操作，一组操作过程转化为矩阵matrix,那么m次操作，即对peanut*(matrix^m)

然后 对题中样例就是

初始化下matrix为单位矩阵
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

输入g 1, 就应该在第1列最后一个加1

1 0 0 0

0 1 0 0

0 0 1 0

1 0 0 1

输入s x y就是把x, y列的值都互换

输入e x,就是把x 列的值都清零

然后用矩阵快速幂就OK了。注意要用__int64 

/*ID: sdj22251PROG: subsetLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define LOCA#define MAXN 500005#define INF 100000000#define eps 1e-7#define L(x) x<<1#define R(x) x<<1|1using namespace std;int n, m;struct wwj{    int r, c;    __int64 mat[105][105];}need, pea;void init(){    memset(pea.mat, 0, sizeof(pea.mat));    memset(need.mat, 0, sizeof(need.mat));    pea.r = 1;    pea.c = n;    need.r = n;    need.c = n;    for(int i = 1; i <= n; i++)    {        pea.mat[1][i] = 0;        need.mat[i][i] = 1;    }    pea.mat[1][n] = 1;}wwj multi(wwj x, wwj y){    wwj t;    int i, j, k;    memset(t.mat, 0, sizeof(t.mat));    t.r = x.r;    t.c = y.c;    for(i = 1; i <= x.r; i++)    {        for(k = 1; k <= x.c; k++)        if(x.mat[i][k])        {            for(j = 1; j <= y.c; j++)            t.mat[i][j] += x.mat[i][k] * y.mat[k][j];        }    }    return t;}int main(){#ifdef LOCAL    freopen("d:/data.in","r",stdin);    freopen("d:/data.out","w",stdout);#endif    int i, j, p, x, y, k;    char s[5];    while(scanf("%d%d%d", &n, &m, &k) != EOF)    {        if(n == 0 && m == 0 && k == 0) break;        n++;        init();        while(k--)        {            scanf("%s", s);            if(s[0] == 'g')            {                scanf("%d", &x);                need.mat[n][x]++;            }            else if(s[0] == 'e')            {                scanf("%d", &x);                for(i = 1; i <= n; i++)                need.mat[i][x] = 0;            }            else if(s[0] == 's')            {                scanf("%d%d", &x, &y);                for(i = 1; i <= n; i++)                swap(need.mat[i][x], need.mat[i][y]);            }        }        while(m)        {            if(m & 1)            {                pea = multi(pea, need);            }            need = multi(need, need);            m = m >> 1;        }        for(i = 1; i < n; i++)        printf("%I64d ", pea.mat[1][i]);        puts("");    }    return 0;}