43 lines
1.1 KiB
C++
43 lines
1.1 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 210;
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const int INF = 0x3f3f3f3f;
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int n, m, k;
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int d[N][N];
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// 算法结束后，d[a][b]表示a到b的最短距离
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void floyd() {
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for (int k = 1; k <= n; k++)
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
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}
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int main() {
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cin >> n >> m >> k;
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// floyd初始化
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memset(d, 0x3f, sizeof d); // 任意两点间距离正无穷
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for (int i = 0; i < N; i++) d[i][i] = 0; // 自己和自己是距离为0的
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// 读入数据
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while (m--) {
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int a, b, c;
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cin >> a >> b >> c;
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d[a][b] = min(d[a][b], c); // 保留最短边.(可能有重边，保留最短边)
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}
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// 调用floyd
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floyd();
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// 处理所有询问
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while (k--) {
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int a, b;
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cin >> a >> b;
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// 由于有负权边存在所以约大过INF/2也很合理
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if (d[a][b] > INF / 2)
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puts("impossible");
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else
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printf("%d\n", d[a][b]);
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}
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return 0;
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} |