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2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
const int INF = 0x3f3f3f3f;
int n, m, k;
int d[N][N];
// 算法结束后d[a][b]表示a到b的最短距离
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;
// floyd初始化
memset(d, 0x3f, sizeof d); // 任意两点间距离正无穷
for (int i = 0; i < N; i++) d[i][i] = 0; // 自己和自己是距离为0的
// 读入数据
while (m--) {
int a, b, c;
cin >> a >> b >> c;
d[a][b] = min(d[a][b], c); // 保留最短边.(可能有重边,保留最短边)
}
// 调用floyd
floyd();
// 处理所有询问
while (k--) {
int a, b;
cin >> a >> b;
// 由于有负权边存在所以约大过INF/2也很合理
if (d[a][b] > INF / 2)
puts("impossible");
else
printf("%d\n", d[a][b]);
}
return 0;
}