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题目描述
In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad's back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley's engagement falls through. Consider Dick's back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.
输入描述:
The first line contains 0 < n <= 100, the number of freckles on Dick's back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.
输出描述:
Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.
输入
3 1.0 1.0 2.0 2.0 2.0 4.0
输出
3.41
C++实现：
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int Tree[101];
int findRoot(int n){
if(Tree[n]==-1)return n;
else{
int tmp=findRoot(Tree[n]);
Tree[n]=tmp;
return tmp;
}
}
struct Point{
float x;
float y;
//Point& operator = (const Point &a)const{
//this->x=a->x;
//this->y=a->y;
}point[101];
struct Edge{
int a;
int b;
float cost;
}edge[4951];
bool cmp(Edge e1,Edge e2){
return e1.cost<e2.cost;
}
float distance(Point a,Point b){
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main(){
int num;
while(scanf("%d",&num)!=EOF){
for(int i=0;i<num;i++){
cin>>point[i].x>>point[i].y;
}
int k=0;
for(int i=0;i<num;i++){
for(int j=i+1;j<num;j++){
//此处为i和j，而不是point结构体，可以在标记根时省力
edge[k].a=i;
edge[k].b=j;
edge[k].cost=distance(point[i],point[j]);
k++;
}
}
sort(edge,edge+num*(num-1)/2,cmp);
for(int i=0;i<num;i++){
Tree[i]=-1;
}
float sum=0;
for(int i=0;i<num*(num-1)/2;i++){
//注意此处findRoot()内不是Tree[edge[i].a/b]
int a=findRoot(edge[i].a);
int b=findRoot(edge[i].b);
if(a!=b){
Tree[a]=b;
sum+=edge[i].cost;
}
}
printf("%.2f\n",sum);
}
}