最大生成ツリー

POJ 2377

           Prim Kruskal  ，                 。             ，                         ，      。

n個のノードを有する連通図の生成ツリーは、原図の極小連通サブ図であり、原図中のすべてのn個のノードを含み、図連通を維持する最小限のエッジを有する.[1]最小生成ツリーは、kruskal(クルーズカール)アルゴリズムまたはprim(プリム)アルゴリズムで求めることができる.
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu
Submit 
Status 
Practice 
POJ 2377
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 
* 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
#include #include #include #include #include using namespace std; const int N = 1005; const int inf = 0x3f3f3f3f; int n, m; int G[N][N]; int dij() {     int dist[N], i, j, v, ans=0;     bool vis[N];     for(i=1; i<=n; i++)     {         vis[i]=0;         dist[i]=G[i][1];     }     vis[1]=1;     for(j=1; j     {         int Max=0;         for(i=1; i<=n; i++)         {             if(!vis[i]&&dist[i]>Max)             {                 Max=dist[i];                 v=i;             }         }         if(Max==0)             return -1;         ans+=Max;         vis[v]=1;         for(i=1; i<=n; i++)         {             if(!vis[i]&&dist[i]                 dist[i]=G[v][i];         }     }     return ans; } int main(){int i,a,b,c;while(~scanf("%d%d",&n,&m)){memset(G,0,sizeof(G));while(m−){scanf("%d%d%%d",&a,&b,&c);if(G[a][b]2はG[a][b]=G[b][a]=cを更新しない;        }         printf("%d",  dij());     }     return 0; }