HDU-1162 Eddy's picture（最小生成树[Kruskal]）

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

   
   
   
   

    
    
    
    3
1.0 1.0
2.0 2.0
2.0 4.0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3.41
   
   
   
   

Kruskal算法就是并查集，学习的同时还能复习并查集

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int INF=0x3f3f3f3f;
const double EPS=0.000001;

struct Edge {
    int s,e,w;
    Edge(int ss=0,int ee=0,int ww=INF):s(ss),e(ee),w(ww) {}
    bool operator < (const Edge& a) const {
        return w<a.w;
    }
}g[10005];

int x[105],y[105],par[10005],pa,pb;

int getPar(int a) {
    if(par[a]==a)
        return a;
    par[a]=getPar(par[a]);
    return par[a];
}

void Merge(int a,int b) {
    pa=getPar(a),pb=getPar(b);
    if(pa!=pb)
        par[pb]=pa;
}

int main() {
    int n,m,i,j,dx,dy,cnt;
    double xx,yy,ans;
    while(scanf("%d",&n)==1) {
        ans=m=0;
        cnt=1;
        for(i=0;i<n;++i) {
            scanf("%lf%lf",&xx,&yy);
            x[i]=int((xx+EPS)*10),y[i]=int((yy+EPS)*10);//防止产生精度问题，坐标乘以10保存为整型
            for(j=0;j<i;++j) {
                dx=x[i]-x[j],dy=y[i]-y[j];
                g[m].s=i,g[m].e=j,g[m].w=dx*dx+dy*dy;//防止产生精度问题，保存两点距离的平方
                ++m;
            }
            par[i]=i;
        }
        sort(g,g+m);
        for(i=0;cnt!=n;++i) {
            if(getPar(g[i].s)!=getPar(g[i].e)) {
                Merge(g[i].s,g[i].e);
                ++cnt;
                ans+=sqrt(double(g[i].w))/10;
            }
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}