AcWing 854. Floyd求最短路
原题链接
简单
作者:
芝居気
,
2021-12-05 19:26:16
,
所有人可见
,
阅读 85
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m, q;
const int N = 250;
int g[N][N];
void floyd()
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++)
{
if (g[j][k] > g[j][i] + g[i][k])
{
g[j][k] = g[j][i] + g[i][k];
}
}
}
}
int main()
{
memset(g, 0x3f, sizeof(g));
cin >> n >> m >> q;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (i == j)
g[i][j] = 0;
for (int i = 0; i < m; i++)
{
int a, b, c;
cin >> a >> b >> c;
g[a][b] = min(g[a][b],c);
}
floyd();
for (int i = 0; i < q; i++)
{
int a, b;
cin >> a >> b;
if (g[a][b] > 0x3f3f3f3f / 2)cout << "impossible" << endl;
else cout << g[a][b] << endl;
}
}