个人心得：这在一定途径上完成查询方面还是很吃力，得多锻炼空间能力，不能再每次都看到就后退，要全力应对，

那怕被虐的不要不要的。

这题主要是求俩个端点中所有路径中最大构成的集合中最小的数值，其实开始思想已经到触及到了这一块，

就想着从一直衍生每次都是max更新，但最终点那里还是卡了一下，还有自己的算法很容易就被突兀的途径给打乱了。

Floyd算法挺不错的，进行一点点的改变就好了，他就是保存每个端点中的最大值的最小值，你想呀，你如果到一个端点，其实在

延伸的时候就已经得到了这条途径的最大值了，比如5端点，你可以从1-3，3-5，则你判断的时候就只要将1-3端点的最大值和3-5端点

的值进行对比就得到了，此时再与5端点所存在的最大值比较就好了

核心算法就是

1 for(int k=1;k<=x;k++)
2             for(int i=1;i<=x;i++)
3                 for(int j=1;j<=x;j++)
4                     if(decibel[i][k]!=inf&&decibel[k][j]!=inf)
5                    decibel[i][j]=min(decibel[i][j],max(decibel[i][k],decibel[k][j]));

Consider yourself lucky! Consider yourself lucky to be still breathing and having fun participating in this contest. But we apprehend that many of your descendants may not have this luxury. For, as you know, we are the dwellers of one of the most polluted cities on earth. Pollution is everywhere, both in the environment and in society and our lack of consciousness is simply aggravating the situation. However, for the time being, we will consider only one type of pollution - the sound pollution. The loudness or intensity level of sound is usually measured in decibels and sound having intensity level 130 decibels or higher is considered painful. The intensity level of normal conversation is 6065 decibels and that of heavy traffic is 7080 decibels. Consider the following city map where the edges refer to streets and the nodes refer to crossings. The integer on each edge is the average intensity level of sound (in decibels) in the corresponding street. To get from crossing A to crossing G you may follow the following path: A-C-F-G. In that case you must be capable of tolerating sound intensity as high as 140 decibels. For the paths A-B-E-G, A-B-D-G and A-C-F-D-G you must tolerate respectively 90, 120 and 80 decibels of sound intensity. There are other paths, too. However, it is clear that A-C-F-D-G is the most comfortable path since it does not demand you to tolerate more than 80 decibels. In this problem, given a city map you are required to determine the minimum sound intensity level you must be able to tolerate in order to get from a given crossing to another. Input The input may contain multiple test cases. The first line of each test case contains three integers C(≤ 100), S(≤ 1000) and Q(≤ 10000) where C indicates the number of crossings (crossings are numbered using distinct integers ranging from 1 to C), S represents the number of streets and Q is the number of queries. Each of the next S lines contains three integers: c1, c2 and d indicating that the average sound intensity level on the street connecting the crossings c1 and c2 (c1 ̸= c2) is d decibels. Each of the next Q lines contains two integers c1 and c2 (c1 ̸= c2) asking for the minimum sound intensity level you must be able to tolerate in order to get from crossing c1 to crossing c2. The input will terminate with three zeros form C, S and Q. Output For each test case in the input first output the test case number (starting from 1) as shown in the sample output. Then for each query in the input print a line giving the minimum sound intensity level (in decibels) you must be able to tolerate in order to get from the first to the second crossing in the query. If there exists no path between them just print the line “no path”. Print a blank line between two consecutive test cases.

Sample Input

7 9 3

1 2 50

1 3 60

2 4 120

2 5 90

3 6 50

4 6 80

4 7 70

5 7 40

6 7 140

1 7

2 6

6 2

7 6 3

1 2 50

1 3 60

2 4 120

3 6 50

4 6 80

5 7 40

7 5

1 7

2 4

0 0 0

Sample Output

Case #1

80 60 60

Case #2

40 no path 80

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iomanip>
 6 #include<algorithm>
 7 using namespace std;
 8 const long long inf=9999999;
 9 int x,y,z;;
10 int decibel[105][105];
11 void init()
12 {
13     for(int i=1;i<=x;i++)
14         for(int j=1;j<=x;j++)
15           if(i==j) decibel[i][j]=0;
16          else decibel[i][j]=inf;
17
18 }
19 int main()
20 {
21      int flag=1;
22     while(cin>>x>>y>>z)
23     {
24
25         if(!x&&!y&&!z) break;
26         init();
27        for(int i=1;i<=y;i++)
28        {
29            int m,n,q;
30            scanf("%d%d%d",&m,&n,&q);
31            if(decibel[m][n]>q)
32                 decibel[m][n]=decibel[n][m]=q;
33
34        }
35        if(flag>1) cout<<endl;
36         for(int k=1;k<=x;k++)
37             for(int i=1;i<=x;i++)
38                 for(int j=1;j<=x;j++)
39                     if(decibel[i][k]!=inf&&decibel[k][j]!=inf)
40                    decibel[i][j]=min(decibel[i][j],max(decibel[i][k],decibel[k][j]));
41                    cout<<"Case #"<<flag++<<endl;
42        for(int i=1;i<=z;i++)
43        {
44            int m,n;
45            scanf("%d%d",&m,&n);
46            if(decibel[m][n]!=inf)
47            cout<<decibel[m][n]<<endl;
48            else
49             cout<<"no path"<<endl;
50
51        }
52     }
53     return 0;
54 }