TOJ-1348 Freckles

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.

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.

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

Output for Sample Input

3.41

Source: Waterloo Local Contest Sep. 23, 2000

最小生成树的模板，用prim算法最简单了吧。

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <math.h>
using namespace std;
const int INF=0x3f3f3f3f;
int n,vn[105];
double lc[105],edge[105][105];
double prim(){
    int i;
    for(i=0;i<n;i++){
        vn[i] = 0;
    }
    for(i=0;i<n;i++){
        lc[i] = INF;
    }
    
    int k = 0;
    vn[0] = -1;//根节点为0号节点,-1表示放入集合vn,非-1为在集合v-vn 
    double ac = 0;
    for(i=0;i<n-1;i++){
        double mc = INF;
        int mv = -1;
        for(int j=0;j<n;j++){
            if(vn[j]!=-1){
                double t = edge[j][k];
                if(t<lc[j]){
                    lc[j] = t;
                    vn[j] = k;//用点k的边更新vn中点到v-vn中各点的最短边 
                }
                if(lc[j]<mc){
                    mc = lc[j];
                    mv = j;//更新vn中点出发的最短边 
                } 
            } 
        }
        ac += mc;
        k =  mv;
        vn[k] = -1;
    } 
    return ac;
}
int main(){
    int i,j,k;
    double x[105],y[105];
    while(cin>>n){
        for(i=0;i<n;i++){
            cin>>x[i]>>y[i];
        } 
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                edge[i][j]=edge[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]) );
            }
        }
        double ans = prim();
        cout<<showpoint<<fixed <<setprecision(2)<<ans<<endl;
    }
    
    return 0;
}