思路：套用Floyd算法思想，d(i, j) = min(d(i,j), max(d(i,k), d(k,j))，就能很方便求得任意两点之间的最小噪音路径。

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 100 + 5;
int d[maxn][maxn];
int main() {
	int C, S, Q, kase = 0;
	while(scanf("%d%d%d", &C, &S, &Q) == 3 && C+S+Q) {
		if(kase++) printf("\n");
		memset(d, inf, sizeof(d));
		int x, y, cost;
		for(int i = 0; i < S; ++i) {
			scanf("%d%d%d", &x, &y, &cost);
			d[x][y] = d[y][x] = cost; //无向图
		}
		//Floyd
		for(int k = 1; k <= C; ++k)
			for(int i = 1; i <= C; ++i)
				for(int j = 1; j <= C; ++j) {
					if(d[i][k] < inf && d[k][j] < inf) {
						int w = max(d[i][k], d[j][k]);
						d[i][j] = min(d[i][j], w);
					}
				}
		printf("Case #%d\n", kase);
		for(int i = 0; i < Q; ++i) {
			scanf("%d%d", &x, &y);
			if(d[x][y] == inf) printf("no path\n");
			else printf("%d\n", d[x][y]);
		}
	}
	return 0;
}

如有不当之处欢迎指出！