hdoj--2122--Ice_cream’s world III(克鲁斯卡尔)

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1450    Accepted Submission(s): 496

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

 

Author

Wiskey

 

Source

HDU 2007-10 Programming Contest_WarmUp

 

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#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int pre[10010];
struct node
{
	int u,v,val;
}p[10010];
int m,n;
void init()
{
	for(int i=0;i<10010;i++)
	pre[i]=i;
}
int cmp(node s1,node s2)
{
	return s1.val<s2.val;
}
int find(int x)
{
	int r=x;
	while(r!=pre[r])
	r=pre[r];
	while(x!=r)
	{
		int t=pre[x];
		pre[x]=r;
		x=t;
	}
	return r;
}
int join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		pre[fx]=fy;
		return 1;
	}
	return 0;
}
int main()
{
	while(cin>>n>>m)
	{
		init();
		for(int i=0;i<m;i++)
		cin>>p[i].u>>p[i].v>>p[i].val;
		sort(p,p+m,cmp);
		int sum=0;
		for(int i=0;i<m;i++)
		{
			if(join(p[i].u,p[i].v))
			sum+=p[i].val;
		}
		int flog=0;
		int gen=0;
		for(int i=0;i<n;i++)
		{
			if(pre[i]==i)
			gen++;
			if(gen>1)
			flog=1;
		}
		if(flog)
		printf("impossible

");
		else
		printf("%d

",sum);
		
	}
	return 0;
}