UVA10048 Floyd最短路径

此题可以直接套上floyd算法的模板，但是要把加法改成max
对于任意一条至少包含两条边的路径，i->j，一定存在k使得i->j
噪音的最高级等于max(d[i][k], d[k][j]),但i->j路径可能并不唯一，
所以还要取一条最小的：d[i][j] = min(d[i][j], max(d[i][k], d[k][j]));

#include <stdio.h>
#include <string.h>
const int INF = 0x3f3f3f3f;
const int N = 105;

int g

, t = 1, que, n, m;

void floyd() {
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
int maxx = g[i][k] < g[k][j] ? g[j][k] : g[i][k];
if (g[i][j] > maxx)
g[i][j] = maxx;
}
}

int main() {
while (scanf("%d%d%d", &n, &m, &que) && n || m || que) {
memset(g, INF, sizeof(g));
int a, b, c;
for (int i = 0; i < m; i++) {
scanf("%d%d%d", &a, &b, &c);
g[a][b] = g[b][a] = c;
}
floyd();
int sta, end;
if (t != 1)
printf("\n");
printf("Case #%d\n", t++);
while (que--) {
scanf("%d%d", &sta, &end);
if (g[sta][end] < INF)
printf("%d\n", g[sta][end]);
else
printf("no path\n");
}
}
return 0;
}