问题 B: Freckles

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本文深入探讨了在平面坐标系中寻找连接多个点并使总连线长度最小的算法——最小生成树问题。通过具体实例，详细讲解了如何利用迪杰斯特拉算法（Dijkstra's algorithm）来解决这一问题，实现对所有‘雀斑’（点）的有效连接，以最少的‘墨水’（线段长度）绘制完整图案。

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

样例输入
3
2723.62 7940.81
8242.67 11395.00
4935.54 6761.32
9
10519.52 11593.66
12102.35 2453.15
7235.61 10010.83
211.13 4283.23
5135.06 1287.85
2337.48 2075.61
6279.72 13928.13
65.79 1677.86
5324.26 125.56
0
样例输出
8199.56
32713.73

在codeup上显示只运行对了50%
在牛课网上通过了

#include<iostream>
#include<math.h>
using namespace std;
const int maxn=110;
const double inf=10000000000;
int n;
struct node{
    double x,y;
}Node[maxn];
double dis[maxn],G[maxn][maxn],sum;
bool vis[maxn];

double distance(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void djikstra(){
    sum=0;
    fill(dis,dis+maxn,inf);
    fill(vis,vis+maxn,false);
    dis[1]=0;
    for(int i=0;i<n;i++){
        int u=-1;
        double min=inf;
        for(int j=1;j<=n;j++){
            if(vis[j]==false && dis[j]<min){
                u=j;
                min=dis[j];
            }
        }
        if(u==-1)
            return;
        vis[u]=true;
        sum+=dis[u];
        for(int k=1;k<=n;k++){
            if(vis[k]==false && G[u][k]!=inf && dis[k]>G[u][k]){
                dis[k]=G[u][k];
            }
        }
    }
}
int main(){
    double a,b;
    while(scanf("%d",&n)!=EOF && n!=0){
        fill(G[0],G[0]+maxn*maxn,inf);
        for(int i=1;i<=n;i++){
            scanf(" %lf %lf",&Node[i].x,&Node[i].y);
        }
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                G[i][j]=G[j][i]=distance(Node[i],Node[j]);
            }
            G[i][i]=0;
        }
        djikstra();
        printf("%.2lf\n",sum);
    }
    return 0;
}