Hdu 2122

Ice_cream’s world III

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

Total Submission(s): 1350    Accepted Submission(s): 449

[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

2 1
0 1 10

4 0

 

[align=left]Sample Output[/align]

10

impossible

       首先要明确使用的是最小生成树的解决方案，其次题目的要求存在无解的情况这里要做一个判断。代码不难，用的是模版

#include<iostream>
#include<cstdio>
#include<algorithm>
#define inf 9999999
using namespace std;
int pot[1001];
int num,ans;
int find(int x)
{
if(pot[x]==x)return x;
pot[x]=find(pot[x]);
return pot[x];
}//找到其所在集合的根结点（代表城市）
struct node{
int a;
int b;
int c;
}road[10001];//保存道路信息
int cmp(const node a,const node b) {
return a.c<b.c;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=0;i<n;i++)
pot[i]=i;

for(int i=0;i<m;i++)
scanf("%d%d%d",&road[i].a,&road[i].b,&road[i].c);

sort(road,road+m,cmp);
num=0;//记录到达的数目
ans=0;//记录路程（必然最短）
for(int i=0;i<m;i++)
{
int x,y;
x=find(road[i].a);
y=find(road[i].b);
if(x!=y)
{
num++;
ans+=road[i].c;
pot[x]=y;
}
}
if(num<n-1)//判断合法
printf("impossible\n");
else
printf("%d\n",ans);

printf("\n");
}
return 0;
}