[AcWing 854] Floyd求最短路

#include<iostream>

using namespace std;
const int N = 210, INF = 1e9;
int n, m, k;
int d[N][N];
void floyd()
{
	for (int k = 1; k <= n; k ++)
		for (int i = 1; i <= n; i ++)
			for (int j = 1; j <= n; j ++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
int main()
{
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i ++) {
		for (int j = 1; j <= n; j ++) {
			if (i == j)		d[i][j] = 0;
			else	d[i][j] = INF;
		}
	}
	while (m --) {
		int a, b, w;
		cin >> a >> b >> w;
		d[a][b] = min(d[a][b], w);
	}
	floyd();
	while (k --) {
		int a, b;
		cin >> a >> b;
		if (d[a][b] > INF / 2)	puts("impossible");
		else	cout << d[a][b] << endl;
	}
	return 0;
}