### SDNU 1370.Freckles 最小生成樹 kruskal算法

1370.Freckles
Time Limit: 1000 MS    Memory Limit: 131072 KB

Description

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.

Input

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.

Output

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

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

后來發現SDNU后面有道拷過來的題正好是當時出的1229，不過范圍沒有改，還是100，用這道題試了一下，kruskal可以過。所以這題用kruskal再貼一個代碼，以示邊少的情況才能用kruskal。

下面AC代碼：

`#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;typedef struct node{    double x,y;}Point;Point p[1005];double Distance(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}struct edge{    int fr,to,nxt;    double w;};int cmp(edge a,edge b){    return a.w<b.w;}int pre[1005],n,head[1005],cnt;edge e[1000005];void add(int fr,int to,double w){    e[cnt].fr=fr;    e[cnt].to=to;    e[cnt].w=w;    e[cnt].nxt=head[fr];    head[fr]=cnt++;}int fin(int x){    if(x==pre[x])        return x;    return pre[x]=fin(pre[x]);}double Kruskal(){    for(int i=1;i<=n;++i)    {        pre[i]=i;    }    sort(e,e+cnt,cmp);    double ans=0;    for(int i=0;i<cnt;++i)    {        int u=fin(e[i].fr);        int v=fin(e[i].to);        if(u!=v)        {            ans+=e[i].w;            pre[u]=v;        }    }    return ans;}int main(){    int i,j;    while(scanf("%d",&n)!=EOF)    {        cnt=0;        memset(e,0,sizeof(e));        memset(pre,0,sizeof(pre));        memset(p,0,sizeof(p));        memset(head,-1,sizeof(head));        for(i=1;i<=n;i++)        {            scanf("%lf%lf",&p[i].x,&p[i].y);        }        for(i=1;i<=n;i++)        {            for(j=i+1;j<=n;j++)            {                double t=Distance(p[i],p[j]);                add(i,j,t);                add(j,i,t);            }        }        printf("%.2f\n",Kruskal());    }    return 0;}`

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