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hdu2122 Ice_cream’s world III(MST)


Ice_cream’s world III


Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1250    Accepted Submission(s): 413


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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最小生成树 kruskal算法

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
int a,b,cost;
}c[10005];
int fa[1005];
void init(int n)
{
for(int i=0;i<n;i++)
fa[i]=i;
}
bool cmp(node x,node y)
{
return x.cost<y.cost;
}
int find(int x)
{
if(fa[x]!=x) fa[x]=find(fa[x]);
return fa[x];
}
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
init(n);
for(int i=0;i<k;i++)
scanf("%d %d %d",&c[i].a,&c[i].b,&c[i].cost);
sort(c,c+k,cmp);
int sum=0;
for(int i=0;i<k;i++)
{
int x=find(c[i].a);
int y=find(c[i].b);
if(x!=y)
sum+=c[i].cost,fa[x]=y;
}
int count=0;
for(int i=0;i<n;i++)
if(fa[i]==i)
count++;
if(count!=1)
printf("impossible\n\n");
else
printf("%d\n\n",sum);
}
return 0;
}



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