Eddy’s pictureTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7428Accepted Submission(s): 3770
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
只需算出一点到其余各点的距离即可。
Prim算法代码：#include <stdio.h>#include <math.h>#include <string.h>#define MAX 110#define INF 1000000000struct Point{double x,y;};double graph[MAX][MAX] ;double prim(int n){bool visited[MAX];memset(visited,false,sizeof(visited));double lowcost[MAX] ;int closest[MAX] ;for(int i = 0 ; i < n ; ++i){lowcost[i] = graph[0][i] ;closest[i] = 0 ;}lowcost[0] = 0.0 ;visited[0] = false ;for(int i = 0 ; i < n ; ++i){int min = INF , index = 0 ;for(int j = 0 ; j < n ; ++j){if(!visited[j] && lowcost[j]<min){index = j ;min = lowcost[j] ;}}visited[index] = true ;for(int j = 0 ; j < n ; ++j){if(!visited[j] && lowcost[j]>graph[index][j]){lowcost[j] = graph[index][j] ;closest[j] = index ;}}}double sum = 0.0 ;for(int i = 1 ; i < n ; ++i){sum += lowcost[i] ;}return sum ;}double distance(const Point &a ,const Point &b){double x = a.x-b.x , y = a.y-b.y ;return sqrt(x*x+y*y);}int main(){Point p[MAX];int n ;while(~scanf("%d",&n)){for(int i = 0 ; i < n ; ++i){scanf("%lf%lf",&p[i].x,&p[i].y) ;}for(int i = 0 ; i < n ; ++i){for(int j = 0 ; j <= i ; ++j){graph[i][j] = graph[j][i] = distance(p[i],p[j]) ;}}double sum = prim(n) ;printf("%.2lf\n",sum) ;}return 0 ;}Kruskal算法代码：#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define MAX 110#define INF 1000000000using namespace std ;struct Point{double x,y;};struct Edge{int x , y ;double w ;}edge[MAX*MAX];bool cmp(const Edge &a ,const Edge &b){return a.w<b.w ;}int f[MAX] ;void init(){for(int i = 0 ; i < MAX ; ++i){f[i] = i ;}}int find(int x){int r = x ;while(r != f[r]){r = f[r] ;}int temp ;while(x != f[x]){temp = f[x] ; f[x] = r ;x = temp ;}return r;}double kruskal(int m ,int n){sort(edge,edge+m,cmp) ;double lowcost[MAX] ;int index = 0 ;init() ;for(int i = 0 ; i < m ; ++i){int fx = find(edge[i].x) , fy = find(edge[i].y) ;if(fx != fy){lowcost[index++] = edge[i].w ;f[fx] = fy ;} }double sum = 0 ;for(int i = 0 ; i < n-1 ; ++i){sum += lowcost[i] ;}return sum ;}double dis(const Point &a ,const Point &b){double x = a.x-b.x , y = a.y-b.y ;return sqrt(x*x+y*y);}int main(){Point p[MAX];int n ;while(~scanf("%d",&n)){for(int i = 0 ; i < n ; ++i){scanf("%lf%lf",&p[i].x,&p[i].y) ;}int index = 0 ;for(int i = 0 ; i < n ; ++i){for(int j = 0 ; j <= i ; ++j){edge[index].x = i , edge[index].y = j ;edge[index].w = dis(p[i],p[j]) ;index++ ;}}double sum = kruskal(index , n) ;printf("%.2lf\n",sum) ;}return 0 ;}与君共勉
，到底通向了什么样的远方呢？
