uva 11551 - Experienced Endeavour(矩阵快速幂)

题目链接：uva 11551 - Experienced Endeavour

题目大意：给定一个序列的变换，求变换r次后各项的值。

解题思路：矩阵快速幂，不解释。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 55;
const int MOD = 1000;

struct Mat {
    int r, c, arr[maxn][maxn];
    Mat (int r = 0, int c = 0) {
        this->r = r;
        this->c = c;
        memset(arr, 0, sizeof(arr));
    }

    Mat operator * (const Mat& u) {
        Mat ret(r, u.c);
        for (int k = 0; k < c; k++) {
            for (int i = 0; i < r; i++)
                for (int j = 0; j < u.c; j++)
                    ret.arr[i][j] = (ret.arr[i][j] + arr[i][k] * u.arr[k][j]) % MOD;
        }
        return ret;
    }
};

Mat pow_mat (Mat ret, Mat x, int n) {
    while (n) {
        if (n&1)
            ret = x * ret;
        x = x * x;
        n >>= 1;
    }
    return ret;
}

int main () {
    int cas, N, K;
    scanf("%d", &cas);

    while (cas--) {
        scanf("%d%d", &N, &K);
        Mat a(N, 1);
        for (int i = 0; i < N; i++) {
            scanf("%d", &a.arr[i][0]);
            a.arr[i][0] %= MOD;
        }

        int x, pos;
        Mat b(N, N);
        for (int i = 0; i < N; i++) {
            scanf("%d", &x);
            for (int j = 0; j < x; j++) {
                scanf("%d", &pos);
                b.arr[i][pos] = 1;
            }
        }

        a = pow_mat(a, b, K);

        printf("%d", a.arr[0][0]);
        for (int i = 1; i < N; i++)
            printf(" %d", a.arr[i][0]);
        printf("\n");
    }
    return 0;
}