UVA10048 - Audiophobia(Floyd,最大值的最小化）

UVA10048 - Audiophobia

题目大意:给定一无向图，每条边都有一个权值，现在给你起点和终点，要求你找出起点到终点途经的边的最大值，要求这个值尽量小，到不了输出no path。

解题思路：在floyd过程中，就可以记录下来。G【i】【j】 = min(G【i】【j】, max(G【i】【k】, G【k】【j】));

代码：

#include 
   
     const int maxn = 105; const int MAXD = 0x7f7f7f7f; int C, S, Q; int G[maxn][maxn]; int min (const int a, const int b) { return a < b ? a : b; } int max(const int a, const int b) { return a > b ? a : b; } void init () { for (int i = 1; i <= C; i++) for (int j = 1; j <= C; j++) G[i][j] = (i == j) ? 0: MAXD; } void Floyd() { for (int k = 1; k <= C; k++) for (int i = 1; i <= C; i++) for (int j = 1; j <= C; j++) { if (G[i][k] == MAXD || G[k][j] == MAXD) continue; G[i][j] = min(G[i][j], max(G[i][k],G[k][j])); } } int main () { int T = 0; while (scanf ("%d %d %d", &C, &S, &Q) && (C, S, Q)) { if (T) printf ("\n"); init(); int c1, c2, d; for (int i = 0; i < S; i++) { scanf("%d %d %d", &c1, &c2, &d); G[c1][c2] = G[c2][c1] = d; } Floyd(); printf ("Case #%d\n", ++T); while (Q--) { scanf ("%d%d", &c1, &c2); if (G[c1][c2] == MAXD) printf ("no path\n"); else printf ("%d\n", G[c1][c2]); } } return 0; }