矩阵快速幂变形

题意：给你一个n×n的邻接矩阵，以及给定整数k。求出图中所有经过边数为k的最短路径以及边数为k的最短路径条数。
思路：边数为k的路径条数直接用矩阵快速幂求得。更新最短路的过程中随即更新即可。

代码：

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ull unsigned long long
#define __ ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

const int maxn = 155 + 10;
const int maxm = 1e4 + 10;
const ll mod = 1e9 + 7;

ll n, m, k;

struct matrix {
    ll a1[maxn][maxn], a2[maxn][maxn];

    matrix operator*(const matrix &x) {
        matrix c;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                c.a1[i][j] = 1e18;
                for (int k = 1; k <= n; ++k) {
                    if (c.a1[i][j] > a1[i][k] + x.a1[k][j]) {
                        c.a1[i][j] = a1[i][k] + x.a1[k][j];
                        c.a2[i][j] = (a2[i][k] * x.a2[k][j]) % mod;
                    } else if (c.a1[i][j] == a1[i][k] + x.a1[k][j]) {
                        c.a2[i][j] = (c.a2[i][j] + a2[i][k] * x.a2[k][j]) % mod;
                    }
                }
            }
        }
        return c;
    }

    matrix &operator=(const matrix &x) {
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= n; ++j) {
                a1[i][j] = x.a1[i][j];
                a2[i][j] = x.a2[i][j];
            }
        }
        return *this;
    }
};

matrix A, ans;

int main() {
    __;
    cin >> n >> m >> k;
    memset(A.a1, 63, sizeof A.a1);
    memset(A.a2, 0, sizeof A.a2);
    for (int i = 1; i <= m; ++i) {
        int x, y, z;
        cin >> x >> y >> z;
        A.a1[x][y] = A.a1[y][x] = z;
        A.a2[x][y] = A.a2[y][x] = 1;
    }
    memset(ans.a1, 0, sizeof ans.a1);
    memset(ans.a2, 0, sizeof ans.a2);
    bool ok = 0;
    while (k) {
        if (k & 1) {
            if (ok)ans = ans * A;
            else {
                ans = A;
                ok = 1;
            }
        }
        A = A * A;
        k >>= 1;
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (ans.a1[i][j] > 1e9) {
                cout << 'X' << ' ' << ans.a2[i][j] << ' ';
            } else cout << ans.a1[i][j] << ' ' << ans.a2[i][j] << ' ';
        }
        cout << endl;
    }
    return 0;
}