HDOJ 2122 Ice_cream’s world III(最小生成树prim算法)

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1223    Accepted Submission(s): 403

Problem Description

ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.

 

Input

Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.

 

Output

If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.

 

Sample Input

   
   
   
   

    
    
    
    2 1 0 1 10 4 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    10 impossible
   
   
   
   

 

 

要考虑自环的情况，wa了两次。   最后一组数据也要输出空行，PE一次。

 

prim算法，代码如下：

 

<span style="font-size:12px;">#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f
int n,map[1010][1010],sum;

void prim()
{
	sum=0;
	int i,j,next,min;
	int lowcost[1010],visit[1010];
	memset(visit,0,sizeof(visit));
	for(i=0;i<n;++i)
	   lowcost[i]=map[1][i];
	visit[1]=1;
	for(i=1;i<n;++i)
	{
		min=INF;
		for(j=0;j<n;++j)
		{
			if(!visit[j]&&min>lowcost[j])
			{
				min=lowcost[j];
				next=j;
			}
		}
		if(min==INF)
		{
			printf("impossible\n\n");
			return ;
		}
		sum+=min;
		visit[next]=1;
		for(j=0;j<n;++j)
		{
			if(!visit[j]&&lowcost[j]>map[next][j])
			    lowcost[j]=map[next][j];
		}   
	}
	printf("%d\n\n",sum);
}

int main()
{
	int m,i,j,x,y,c;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				if(i==j)
				   map[i][j]=0;//考虑自环的情况 
				else
				   map[i][j]=INF;
			}
		}
		while(m--)
		{
			scanf("%d%d%d",&x,&y,&c);
			if(map[x][y]>c)//自环的情况不计入 
			    map[x][y]=map[y][x]=c;
		}
		prim();	
	}
	return 0;
}
 </span>