hdu 5137 去掉一个点 使得最短路最大（2014广州区域赛K题）

题意：从2~n-1这几个点中任意去掉一个点，使得从1到n的最短路径最大，如果任意去掉一个点1~n无通路输出Inf。

Sample Input
4 5
1 2 3
1 3 7
1 4 50
2 3 4
3 4 2
3 2
1 2 30
2 3 10
0 0

Sample Output
50
Inf

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <string>
 6 # include <map>
 7 # include <cmath>
 8 # include <queue>
 9 # include <list>
10 # define LL long long
11 using namespace std ;
12
13 const int MAXN=300;
14 const int INF=0x3f3f3f3f;
15 int n  , m ;
16 bool vis[MAXN];
17 int cost[MAXN][MAXN] ;
18 int lowcost[MAXN] ;
19
20 void Dijkstra(int beg)
21 {
22     for(int i=0;i<n;i++)
23     {
24         lowcost[i]=INF;
25     }
26     lowcost[beg]=0;
27     for(int j=0;j<n;j++)
28     {
29         int k=-1;
30         int Min=INF;
31         for(int i=0;i<n;i++)
32             if(!vis[i]&&lowcost[i]<Min)
33             {
34                 Min=lowcost[i];
35                 k=i;
36             }
37             if(k==-1)
38                 break ;
39             vis[k]=true;
40             for(int i=0;i<n;i++)
41                 if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i])
42                 {
43                     lowcost[i]=lowcost[k]+cost[k][i];
44                 }
45     }
46
47 }
48
49 int main ()
50 {
51     //freopen("in.txt","r",stdin) ;
52     while (scanf("%d %d" , &n , &m) !=EOF)
53     {
54         if (n == 0 && m == 0)
55             break ;
56
57         int u , v , w ;
58         int i , j ;
59         int MAX = 0 ;
60         for (i = 0 ; i < n ; i++)
61         for (j = 0 ; j < n ; j++)
62        {
63            if (i == j)
64               cost[i][j] = 0 ;
65            else
66               cost[i][j] = INF ;
67        }
68         while(m--)
69         {
70             scanf("%d%d%d" , &u , &v , &w) ;
71             cost[u-1][v-1] = w ;
72             cost[v-1][u-1] = w ;
73
74         }
75
76         for (i = 1 ; i <= n-2 ; i++)
77         {
78             memset(vis , 0 , sizeof(vis)) ;
79             vis[i] = 1 ;
80             Dijkstra(0) ;
81             if (lowcost[n-1] > MAX )
82                 MAX = lowcost[n-1] ;
83
84         }
85
86         if (MAX == INF)
87            printf("Inf\n") ;
88         else
89            printf("%d\n" , MAX) ;
90     }
91
92     return 0 ;
93 }