POJ 3613 Cow Relays（矩阵模板+快速幂+floyd）

题目地址
题意：给你一个图，要求你一定要经过K步从s到达t，求最短路为多少
思路：通过矩阵快速幂以及floyd的思想定义出一个矩阵，一个矩阵代表的是经过了n条边从i到达j，另一个矩阵代表的是经过了m条边从i到达j，则两个矩阵相乘代表的是经过了n+m条边从i到达j，这样就能求出来了。
在知乎看到了为什么floyd要把k放在最外层：

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#define N 210
#define LL __int64
#define inf 0x3f3f3f3f
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
#define getMid (l+r)>>1
#define movel ans<<1
#define mover ans<<1|1
using namespace std;
const LL mod = 1000000007;
int K, n, m, s, t;
struct Matrix {
    int a[N][N];
    Matrix operator * (Matrix &r) {
        Matrix c;
        memset(c.a, inf, sizeof(c.a));
        for (int i = 1; i <= n; i++)//floyd不同的是把k放在内层了
            for (int j = 1; j <= n; j++)
                for (int k = 1; k <= n; k++)
                    c.a[i][j] = min(c.a[i][j], a[i][k] + r.a[k][j]);
        return c;
    }
}ans, cnt;
map<int, int> mapp;
void power() {
    cnt = ans;
    K--;
    while (K) {
        if (K & 1) ans = ans*cnt;
        cnt = cnt * cnt;
        K >>= 1;
    }
}
int main() {
    cin.sync_with_stdio(false);
    int a, b, c;
    while (cin >> K >> m >> s >> t) {
        memset(ans.a, inf, sizeof(ans.a));
        n = 0;
        mapp.clear();
        for (int i = 0; i < m; i++) {
            cin >> c >> a >> b;
            if (mapp.find(a) != mapp.end()) {
                a = mapp[a];
            }
            else {
                mapp[a] = ++n;
                a = n;
            }
            if (mapp.find(b) != mapp.end()) {
                b = mapp[b];
            }
            else {
                mapp[b] = ++n;
                b = n;
            }
            ans.a[a][b] = ans.a[b][a] = c;
        }
        power();
        cout << ans.a[mapp[s]][mapp[t]] << endl;
    }
    return 0;
}