POJ 3270. Cow Sorting & 51nod 1125 交换机器的最小代价

根据题意，需要交换的部分会形成若干个不相交的环，独立处理每个环。

每个环可以用环内的最小值去和其它元素交换，或者用全局最小值和环上最小值交换，做一遍再交换回去。

#include <cstdio>
#include <cstring>

const int MOD = 9973;
int m, n, k;

void M(int &a) {
    if (a >= MOD) a -= MOD;
    if (a < 0) a += MOD;
}

struct Mat {
    int mat[11][11];
    void clear() {
        memset(mat, 0, sizeof(mat));
    }
    Mat(int x = 0) {
        memset(mat, 0, sizeof(mat));
        for (int i = 1; i <= m; i++)
            mat[i][i] = x;
    }
    Mat operator * (const Mat &p) const {
        Mat c;
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= m; j++) {
                int cnt = 0;
                for (int k = 1; k <= m; k++)
                    cnt += mat[i][k] * p.mat[k][j];
                c.mat[i][j] = cnt % MOD;
            }
        return c;
    }
    Mat operator ^ (int b) {
        Mat c(1);
        Mat a = *this;
        while (b) {
            if (b & 1) c = c * a;
            a = a * a;
            b >>= 1;
        }
        return c;
    }
} base, res;

int phi(int n) {
    int ans = n;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            ans -= ans / i;
            while (n % i == 0)
                n /= i;
        }
    }
    if (n > 1)
        ans -= ans / n;
    return ans % MOD;
}

int qp(int a, int b = MOD - 2) {
    int ans = 1;
    a %= MOD;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

int solve(int x) {
    res = base ^ x;
    int ans = 0;
    for (int i = 1; i <= m; i++)
        M(ans += res.mat[i][i]);
    return ans;
}

int solve() {
    int ans = 0;
    for (int i = 1; i * i <= n; i++) {
        if (n % i) continue;
        M(ans += 1LL * phi(n / i) * solve(i) % MOD);
        if (i * i == n) continue;
        M(ans += 1LL * phi(i) * solve(n / i) % MOD);
    }
    return ans * qp(n) % MOD;
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        base.clear();
        scanf("%d%d%d", &n, &m, &k);
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= m; j++)
                base.mat[i][j] = 1;
        while (k--) {
            int a, b;
            scanf("%d%d", &a, &b);
            base.mat[a][b] = base.mat[b][a] = 0;
        }
        printf("%d\n", solve());
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Mrzdtz220/p/12231022.html