poj 2485 求最小生成树中 最大的一个权值

Sample Input

1 //T

3 //n
0 990 692 //邻接矩阵
990 0 179
692 179 0
Sample Output

692

prim

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <cmath>
 6 # define LL long long
 7 using namespace std ;
 8
 9 const int INF=0x3f3f3f3f;
10 const int MAXN=510;
11 bool vis[MAXN];
12 int lowc[MAXN];
13 int n ;
14 int cost[MAXN][MAXN] ;
15 int MAX = 0 ;
16
17 int Prim()//点是0~n-1
18 {
19     int ans=0;
20     memset(vis,false,sizeof(vis));
21     vis[0]=true;
22     for(int i=1;i<n;i++)lowc[i]=cost[0][i];
23     for(int i=1;i<n;i++)
24     {
25         int minc=INF;
26         int p=-1;
27         for(int j=0;j<n;j++)
28             if(!vis[j]&&minc>lowc[j])
29             {
30                 minc=lowc[j];
31                 p=j;
32             }
33             if(minc==INF)return -1;//原图不连通
34             ans+=minc;
35             if (minc > MAX)
36                 MAX = minc ;
37             vis[p]=true;
38             for(int j=0;j<n;j++)
39                 if(!vis[j]&&lowc[j]>cost[p][j])
40                     lowc[j]=cost[p][j];
41     }
42     return ans;
43 }
44
45 int main()
46 {
47
48    // freopen("in.txt","r",stdin) ;
49     int T ;
50     cin>>T ;
51     while(T--)
52     {
53         cin>>n ;
54         int w ;
55         int i , j ;
56         for (i = 0 ; i < n ; i++)
57             for (j = 0 ; j < n ; j++)
58                {
59                    cin>>w ;
60                    if(w==0)
61                      cost[i][j] = INF ;
62                    else
63                      cost[i][j] = w ;
64                }
65         MAX = 0 ;
66         Prim() ;
67         cout<<MAX<<endl ;
68
69     }
70     return 0 ;
71 }