杭电2122Ice_cream’s world III

Ice_cream’s world III

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

Total Submission(s): 1356    Accepted Submission(s): 453


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

 
好久没写代码了，挺简单的一道题，不知道为啥么没人做，就是最基础的最小生成树，这里用的最小生成树的库鲁斯卡尔算法，该算法用的并查集，很基础很重要的算法 ；
附ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int start;
int end;
int cost;
}t[100100];
int per[10001];
int cmp(node a,node b)
{
return a.cost<b.cost;
}
int find(int x)
{
int r=x;
while(r!=per[r])
r=per[r];
return r;
}
int join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
per[fx]=fy;
return 1;
}
return 0;
}
int main()
{
int i,j,k,l,m,n;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=0;i<10001;i++)
per[i]=i;
for(i=0;i<n;i++)
scanf("%d%d%d",&t[i].start,&t[i].end,&t[i].cost);
sort(t,t+n,cmp);
int ans=0;
for(i=0;i<n;i++)
{
if(join(t[i].start,t[i].end))
ans+=t[i].cost;
}
int flag=0;
for(i=0;i<m;i++)
if(per[i]==i)
flag++;
if(flag>1)
printf("impossible\n\n");
else
printf("%d\n\n",ans);
}
return 0;
}