杭电 最小生成树 2122 Ice_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): 1446    Accepted Submission(s): 494

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 10 1 104 0

 

Sample Output

10impossible

 

这个题目的要求和 Ice_cream’s world II一样，要求遍历所有的点。另外要求保证只有一条道路。

看到了遍历所有的点，我们直接就想到了最小生成树。直接上代码即可：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;struct dian{    int x,y,dis;}a[121212];int cmp(dian a,dian b){    return a.dis<b.dis;}int f[121212];int find(int a){    int r=a;    while(f[r]!=r)    r=f[r];    int i=a;    int j;    while(i!=r)    {        j=f[i];        f[i]=r;        i=j;    }    return r;}int merge(int a,int b){    int A,B;    A=find(a);    B=find(b);    if(A!=B)    f[B]=A;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        memset(f,0,sizeof(f));        for(int i=0;i<n;i++)        {            f[i]=i;        }        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].dis);        }        sort(a,a+m,cmp);        int output=0;        for(int i=0;i<m;i++)        {            if(find(a[i].x)!=find(a[i].y))            {                merge(a[i].x,a[i].y);                output+=a[i].dis;            }        }        int flag=0;        for(int i=0;i<n;i++)        {            //printf("%d ",f[i]);            if(f[i]==i)flag++;        }        printf("\n");        if(flag==1)        {            printf("%d\n\n",output);        }        else            printf("impossible\n\n");    }}