luogu P2886 [USACO07NOV]Cow Relays G Floyd算法求从S到E恰好经过K条边的最短路

#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
const int N = 1010;
int k, n, m, S, E;
//距离
int g[N][N];
int res[N][N];
void mul(int c[][N], int a[][N], int b[][N])
{
	static int temp[N][N];
	memset(temp, 0x3f, sizeof temp);
	for (int k = 1; k <= n; k ++ )
		for (int i = 1; i <= n; i ++ )
			for (int j = 1; j <= n; j ++ )
				temp[i][j] = min(temp[i][j], a[i][k] + b[k][j]);
	memcpy(c, temp, sizeof temp);
}
void qmi()
{
	memset(res, 0x3f, sizeof res);
	for (int i = 1; i <= n; i ++ )
		res[i][i] = 0;
	while (k)
	{
		if (k & 1)
			mul(res, res, g);    // res = res * g
		mul(g, g, g);   // g = g * g
		k >>= 1;
	}
}
int main()
{
	cin >> k >> m >> S >> E;
	memset(g, 0x3f, sizeof g);
	//离散化
	map<int, int> ids;
	if (!ids.count(S))
		ids[S] = ++ n;
	if (!ids.count(E))
		ids[E] = ++ n;
	S = ids[S], E = ids[E];
	while (m -- )
	{
		int a, b, c;
		cin >> c >> a >> b;
		if (!ids.count(a))
			ids[a] = ++ n;
		if (!ids.count(b))
			ids[b] = ++ n;
		a = ids[a], b = ids[b];
		g[a][b] = g[b][a] = min(g[a][b], c);
	}
	qmi();
	cout << res[S][E] << endl;
	return 0;
}