题目大意：从a城市到b城市的路径中，尽可能让一路上的最大噪音最小。
题目思路：设d [ i ][ j ]表示 i 到 j 的最大噪音的最小值。 那么d [ i ][ j ] = min( d[ i ][ j ] ,max( d [ i ][ k ] , d [ k ][ j ]) ); AC代码：
代码语言：javascript复制#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<set>
#include<utility>
#include<map>
#include<string>
using namespace std;
const int maxn = 100 100;
const int INF = 0x3f3f3f3f;
int C, S, Q;
int c1, c2, d;
int dist[maxn][maxn];
void Floyd()
{
for(int k = 1; k <= C; k )
{
for(int i = 1; i <= C; i )
for(int j = 1; j <= C; j )
dist[i][j] = min(dist[i][j], max(dist[i][k], dist[k][j]));
}
}
int main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int cnt = 0;
while(cin >> C >> S >> Q && C && S && Q)
{
cnt;
for(int i = 1; i <= C; i )
for(int j = 1; j <= C; j )
dist[i][j] = INF;
for(int i = 0; i < S; i )
{
cin >> c1 >> c2 >> d;
dist[c1][c2] = dist[c2][c1] = d;
}
Floyd();
if(cnt > 1)
cout << endl;
cout << "Case #" << cnt << endl;
for(int i = 0; i < Q; i )
{
cin >> c1 >> c2;
if(dist[c1][c2] == INF)
cout << "no path" << endl;
else
cout << dist[c1][c2] << endl;
}
}
}
