杭电 3371 kruskal（）算法

           话说这道蛋疼的题，是专题里面的，我现在还是没有过，，，，，，TLE到吐血，，TLE了20多次，，，还是过不了。G++真让人蛋疼，，，在外面用C++，600ms过了。蛋疼，，，，，，，听说用prime900多ms可以过，，囧，，，明天看看prime算法，，再试试。。。。。题目：

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2756    Accepted Submission(s): 844

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  

 

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

 

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

 

Sample Input

   
   
   
   

    
    
    
    1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
   
   
   
   

 

#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=25010;
int leftt[N],rightt[N],value[N],r[N],father[N];
int cmp(int i,int j){
  return value[i]<value[j];
}
int find(int x){
  if(father[x]!=x)
      father[x]=find(father[x]);
  return father[x];
}
int main(){
  int kk;
  scanf("%d",&kk);
  while(kk--){
      for(int i=0;i<N;++i){
        leftt[i]=0;rightt[i]=0;
        value[i]=1000005;
        r[i]=i;
        father[i]=i;
      }
      int n,m,k;
      scanf("%d%d%d",&n,&m,&k);
      int a,b,c,i,j,num=0;
      for(i=0;i<m;++i){
        scanf("%d%d%d",&leftt[i],&rightt[i],&value[i]);
       }
      int t,x,y;
      while(k--){
        scanf("%d",&t);
        scanf("%d",&x);
        t--;
        while(t--){
          scanf("%d",&y);
          leftt[m]=x;
          rightt[m]=y;
          father[m]=x;
          value[m]=0;
          m++;
        }
      }
      /*for(i=0;i<m;++i)
          printf("%d ",father[i]);
      printf("\n");*/
      sort(r,r+m,cmp);
      int sum=0;
      for(i=0;i<m;++i){
        int pos=r[i];
        /*printf("%d\n",pos);*/
        int xx=find(leftt[pos]);
        int yy=find(rightt[pos]);
        /*printf("xx=%d   yy=%d\n",xx,yy);*/
        if(xx!=yy){
          father[yy]=xx;
          sum+=value[pos];
         /* printf("sum=%d\n",sum);*/
        }
      }
      int flag=1;
      int zz=find(father[1]);
      for(i=2;i<=n;++i){
        if(zz!=father[i]&&zz!=find(father[i]))
        {flag=0;break;}
      }
      if(flag)
      printf("%d\n",sum);
      else
          printf("-1\n");
  }
  return 0;
}