poj 2377 Bad Cowtractors （最大生成树prim）

Bad Cowtractors

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Problem Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to
do the worst job possible. She must decide on a set of connections to install so
that (i) the total cost of these connections is as large as possible, (ii) all
the barns are connected together (so that it is possible to reach any barn from
any other barn via a path of installed connections), and (iii) so that there are
no cycles among the connections (which Farmer John would easily be able to
detect). Conditions (ii) and (iii) ensure that the final set of connections will
look like a "tree".

 

Input

* Line 1: Two space-separated integers: N and M
<br> <br>* Lines 2..M+1: Each line contains three space-separated
integers A, B, and C that describe a connection route between barns A and B of
cost C.

 

Output

* Line 1: A single integer, containing the price of the
most expensive tree connecting all the barns. If it is not possible to connect
all the barns, output -1.

 

Sample Input

5 8
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17

 

Sample Output

42

 #include <iostream>
 #include <cstdio>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 int a[][];
 int dis[];
 bool vis[];
 int n, m;
 void Prime()
 {
     for (int i = ; i <= n; i++)
     {
         vis[i] = false;
         dis[i] = a[][i];
     }
     dis[] = ;
     vis[] = true;
     int ans = ;
     for (int i = ; i <= n; i++)
     {
         int minn = ;
         int p = -;
         for (int j = ; j <= n; j++)
         {
             if (!vis[j] && dis[j]>minn)// 是大于，找出最大的边
                 minn = dis[p = j];
         }
         if (p == -)
         {
             cout << "-1" << endl;
             return;
         }
         vis[p] = true;
         ans += minn;
         for (int j = ; j <= n; j++)
         {
             if (!vis[j] && dis[j]<a[p][j])//尽可能让边变大
                 dis[j] = a[p][j];
         }
     }
     cout << ans << endl;
 }
 int main()
 {
     while (cin >> n >> m)
     {
         //初始化为0
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j <= n; j++)
                 a[i][j] = ;
         }
         int x, y, z;
         while (m--)
         {
             scanf("%d%d%d", &x, &y, &z);
             if (z>a[x][y])
                 a[x][y] = a[y][x] = z;
         }
         Prime();
     }
     return ;
 }