HDU 3371（最小生成树，Prim）

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1271    Accepted Submission(s): 423

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 <stdio.h>
#include <string.h>
#define MAX 505
#define MAXCOST 0x7fffffff

int sum,flag,num;


void Prim(int graph[][MAX],int n)   
{   
    int i,j,k,min;   
    int lowcost[MAX];   
    int adjvex[MAX];   
    for(i=1;i<n;i++)   
        lowcost[i]=graph[0][i];   //从第一个的顶点开始   
    memset(adjvex,0,sizeof(adjvex));     
    min=MAXCOST;   
    for(i=1;i<n;i++)   
    {   
        min=MAXCOST;   
        for(j=1;j<n;j++)   
            if(adjvex[j]==false && lowcost[j]<min)   
            {   
                min=lowcost[j];   
                k=j;    //记下最小的点   
            }   
        adjvex[k]=true;   //为true表示该权值已经是最小，为flase是还不确定，应继续更新   
        for(j=1;j<n;j++)   
        {   
            if(adjvex[j]==false && lowcost[j]>graph[k][j])  //更新lowcost   
                lowcost[j]=graph[k][j];   
        }   
    }   
    for(i=1;i<n;i++)  
	{
		if(lowcost[i]!=MAXCOST)
		{
			sum+=lowcost[i];  //这就是最小生成树   
			num++;
		}
	}
} 

int main()
{
	int n,m,i,j,x,y,a,b,c;
	int graph[MAX][MAX],q[MAX];
	scanf("%d",&m);
	while (m--)
	{
		flag=0;
		num=0;
		scanf("%d%d%d",&n,&x,&y);   //顶点个数
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				graph[i][j]=MAXCOST;
		while(x--)
		{
			scanf("%d%d%d",&a,&b,&c);
			if(graph[a-1][b-1]>c)
				graph[a-1][b-1]=graph[b-1][a-1]=c;
		}
		while(y--)
		{
			int l,k,p;
			scanf("%d",&p);
			for(l=0;l<p;l++)
			{
				scanf("%d",&q[l]);
				q[l]--;
			}
			for(k=0;k<p;k++)
				for(l=0;l<p;l++)
					graph[q[k]][q[l]]=graph[q[l]][q[k]]=0;
		}
		sum=0;
		Prim(graph,n);
		if(num==n-1)
			flag=1;
		if(flag)
			printf("%d/n",sum);
		else
			printf("-1/n");
	}
	return 0;
}