HDU 2122 Ice_cream’s world III【最小生成树】

解题思路：基础的最小生成树
反思：不明白为什么i从1开始取，就一直WA,难道是因为村庄的编号是从0开始的吗

Ice_cream’s world III

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1032    Accepted Submission(s): 335

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int pre[10010];
struct Edge
{

    int u,v,w;
} edge[10010];
bool cmp(Edge n1,Edge n2)
{
   return n1.w<n2.w;
}
int find(int root)
{
   return root == pre[root] ? root : pre[root] = find(pre[root]);
}
int unionroot(int x,int y)
{
	int root1=find(x);
	int root2=find(y);
	if(root1==root2)
	return 0;
	pre[root1]=root2;
	return 1;
}
int main()
{
    int n,m,i,j,tmp,x,y,ans;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
    	tmp=0;
    	ans=0;
        for(i=0;i<=10010;i++)
        pre[i]=i;
        for(i=0;i<m;i++)
        scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].w);

        sort(edge,edge+m,cmp);
        for(i=0;i<m;i++)
        {
        	x=find(edge[i].u);
        	y=find(edge[i].v);
        	if(unionroot(x,y))
        	{
        		ans+=edge[i].w;
        		tmp++;
        	}
        }
        if(tmp!=n-1)
        printf("impossible\n");
        else
        printf("%d\n",ans);
		printf("\n");
    }
}