poj2237 Bad Cowtractors 最大生成树

D - Bad Cowtractors

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

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 81 2 31 3 72 3 102 4 42 5 83 4 63 5 24 5 17

Sample Output

42

Hint

OUTPUT DETAILS: 

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

题意就是让我们求最大生成树

思路:在这里我用的Kruskal算法,对边的权值按照从大到小排序,依次选取权值最大的边，然后用并查集判断是否连通,直到选取了
n-1条变时结束；

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{   int u;
    int v;
    int w;
};
int n,m;
int pre[1100];
int cmp(node a,node b)
{    return a.w>b.w;
}
int find(int x)
{    if(x==pre[x])
      return x;
     else
      {
    pre[x]=find(pre[x]);
    return pre[x];}
}
int join(int x1,int x2)
{    int f1=find(x1);
     int f2=find(x2);
     if(f1!=f2)
      {
  pre[f2]=f1;
      return  1;}
     return 0;
}
int main()
{    int i,j;
     scanf("%d%d",&n,&m);
     int flag=0;
     //Kruskal算法核心; 
       struct node q[20010];
         for(i=1;i<=n;i++)
          pre[i]=i;
         for(i=1;i<=m;i++)
          scanf("%d %d %d",&q[i].u,&q[i].v,&q[i].w);
          sort(q+1,q+m+1,cmp);
           int sum=0,count=0;
           for(i=1;i<=m;i++)
           {   if(join(q[i].u,q[i].v))
                {  count++;
                   sum+=q[i].w;
}
if(count==n-1)
{  flag=1;
   break;}
  }
  if(flag)
  printf("%d\n",sum);
  else
  printf("-1");
return 0;
}