Swordfish （Kruskal）

  We all remember that in the movie Swordfish, Gabriel broke into the World Bank Investors Group in West Los Angeles, to rob $9.5 billion. And he needed Stanley, the best hacker in the world, to help him break into the password protecting the bank system. Stanley's lovely daughter Holly was seized by Gabriel, so he had to work for him. But at the last moment, Stanley made some little trick in his hacker mission: he injected a trojan horse in the bank system, so the money would jump from one account to another account every 60 seconds, and would continue jumping in the next 10 years. Only Stanley knew when and where to get the money. If Gabriel killed Stanley, he would never get a single dollar. Stanley wanted Gabriel to release all these hostages and he would help him to find the money back.
  You who has watched the movie know that Gabriel at last got the money by threatening to hang Ginger to death. Why not Gabriel go get the money himself? Because these money keep jumping, and these accounts are scattered in different cities. In order to gather up these money Gabriel would need to build money transfering tunnels to connect all these cities. Surely it will be really expensive to construct such a transfering tunnel, so Gabriel wants to find out the minimal total length of the tunnel required to connect all these cites. Now he asks you to write a computer program to find out the minimal length. Since Gabriel will get caught at the end of it anyway, so you can go ahead and write the program without feeling guilty about helping a criminal.

Input:
The input contains several test cases. Each test case begins with a line contains only one integer N (0 <= N <=100), which indicates the number of cities you have to connect. The next N lines each contains two real numbers X and Y(-10000 <= X,Y <= 10000), which are the citie's Cartesian coordinates (to make the problem simple, we can assume that we live in a flat world). The input is terminated by a case with N=0 and you must not print any output for this case.

Output:
You need to help Gabriel calculate the minimal length of tunnel needed to connect all these cites. You can saftly assume that such a tunnel can be built directly from one city to another. For each of the input cases, the output shall consist of two lines: the first line contains "Case #n:", where n is the case number (starting from 1); and the next line contains "The minimal distance is: d", where d is the minimal distance, rounded to 2 decimal places. Output a blank line between two test cases.

Sample Input:

50 00 11 11 00.5 0.50

Sample Output:

Case #1:The minimal distance is: 2.83题意：第一行给出点的个数以下N行分别为每个点的横坐标和纵坐标把所有点连在一起，使得距离和最小很明显是最小生成树的问题，但要注意格式（在每个样例前空一行）代码：
#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;int n;struct node2{         double x,y;}num[1000];   //存放坐标struct node1{    int u,v;    double w;}e[100010];    //存放边关系int pre[1000];void init(){    int i;    for(i=0;i<=n;i++)        pre[i]=i;}int Find(int x){    int r=x;    while(r!=pre[r]) //寻找根结点        r=pre[r];    int i=x,j;    while(i!=r){   //压缩路径        j=pre[i];        pre[i]=r;        i=j;    }    return r;}int mix(int a,int b){    int fa=Find(a),fb=Find(b);    if(fa!=fb){        pre[fb]=fa;        return 1;    }    return 0;}int cmp(node1 a,node1 b){    return a.w<b.w;}int main(){    int cnt=0;    while(scanf("%d",&n)!=EOF&&n){        if(cnt) printf("\n");  //注意格式，我在这一直错，无语了......，找了好长时间的格式错误        int i,j;        int t=0;        for(i=1;i<=n;i++){            scanf("%lf%lf",&num[i].x,&num[i].y);                for(j=1;j<=i;j++){                           //求新点与以前点的距离，                 double a=(num[i].x-num[j].x)*(num[i].x-num[j].x);                 double b=(num[i].y-num[j].y)*(num[i].y-num[j].y);                 double s=sqrt(a+b);                 if(s!=0){  //如果不是本身则把边关系存放                    e[t].u=i;                 e[t].v=j;                 e[t++].w=s;                // printf("S  %.3f\n",s);                 }            }        }        init();        sort(e,e+t,cmp);        double sum=0.0;        int countt=0;        for(i=0;i<t;i++){            if(mix(e[i].u,e[i].v)){                 sum+=e[i].w;                countt++;            }            if(countt==n-1) break;        }      printf("Case #%d:\nThe minimal distance is: %.2f\n", ++cnt, sum);    }return 0;}