杭电1162Eddy's picture（最小生成树）

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8107    Accepted Submission(s): 4106

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you.
Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

 

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point. 

Input contains multiple test cases. Process to the end of file.

 

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points. 

 

Sample Input

31.0 1.02.0 2.02.0 4.0

 

Sample Output

3.41

 

又一道模板题，最小生成树的克鲁斯卡尔算法，只是普利姆算法还不是很会用，就一直用克鲁斯卡尔了，不过普利姆算法可以处理大数据，这一点想必是极好的：

附ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;struct node {int start;int end;double dis;}t[10010];int cmp(node a,node b){return a.dis <b.dis ;}int per[110];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 m,n,i,j,p;double x[110],y[110];while(scanf("%d",&n)!=EOF){for(i=0;i<110;i++)per[i]=i;for(i=0;i<n;i++)scanf("%lf%lf",&x[i],&y[i]);int k=0;for(i=0;i<n;i++)for(j=i+1;j<n;j++){t[k].start=i;t[k].end =j;t[k].dis =sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));k++;}sort(t,t+k,cmp);double sum=0.00;for(i=0;i<k;i++)if(join(t[i].start,t[i].end))sum=sum+t[i].dis;printf("%.2lf\n",sum);}return 0;}