题解报告：hdu 1162 Eddy's picture

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?

译文：埃迪最近开始喜欢画画，他肯定自己要成为一名画家。埃迪每天都在他的小房间里画画，他通常会拍出他最新的照片让他的朋友们欣赏。但结果可想而知，朋友对他的照片不感兴趣。埃迪感到非常困惑，为了将所有朋友的观点转变为他的绘画技术，埃迪为他的朋友们创造了一个问题。问题描述如下：给出你在绘图纸上的一些坐标信息，每一点用直线与油墨连接，使所有点最终连接在同一个地方。你有多少距离发现了墨水吸取的最短长度？

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.

译文：第一行包含0 <n <= 100，即点数。对于每一点，一条线跟随; 每个以下行包含两个实数，指示该点的（x，y）坐标。输入包含多个测试用例。处理到文件的结尾。

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. 

译文：你的程序打印一个单一的实数到小数点后两位：可以连接所有点的墨水线的最小总长度。

Sample Input

3

1.0 1.0

2.0 2.0

2.0 4.0

Sample Output

3.41

解题思路：最小生成树，简单题。

AC代码之Prim算法：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n;
bool vis[maxn];
double lowdist[maxn],dist[maxn][maxn];
pair<double,double> point[maxn];
double vecx(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
double Prim(){
    for(int i=1;i<=n;++i)
        lowdist[i]=dist[1][i];
    lowdist[1]=0;vis[1]=true;
    double res=0.0;
    for(int i=1;i<n;++i){
        int k=-1;
        for(int j=1;j<=n;++j)
            if(!vis[j] && (k==-1||lowdist[k]>lowdist[j]))k=j;
        if(k==-1)break;
        vis[k]=true;
        res+=lowdist[k];
        for(int j=1;j<=n;++j)
            if(!vis[j])lowdist[j]=min(lowdist[j],dist[k][j]);
    }
    return res;
}
int main()
{
    while(cin>>n){
        for(int i=1;i<=n;++i)
            cin>>point[i].first>>point[i].second;
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                dist[i][j]=vecx(point[j].first,point[j].second,point[i].first,point[i].second);
        memset(vis,false,sizeof(vis));
        cout<<setiosflags(ios::fixed)<<setprecision(2)<<Prim()<<endl;
    }
    return 0;
}

 AC代码之Kruskal算法：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
const int maxc = 10005;//100*100+5
int n,k,father[maxn];
double lowdist;
pair<double,double> point[maxn];//点的坐标
struct edge{int u,v;double dist;}es[maxc];
bool cmp(const edge& e1,const edge& e2){return e1.dist<e2.dist;}
double vecx(double x1,double y1,double x2,double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int find_father(int x){//找根节点
    int pir=x,tmp;
    while(father[pir]!=pir)pir=father[pir];
    while(x!=pir){
        tmp=father[x];
        father[x]=pir;//路径压缩
        x=tmp;
    }
    return x;
}
void unite_father(int x,int y,double z){
    x=find_father(x);
    y=find_father(y);
    if(x!=y){
        lowdist+=z;
        father[x]=y;
    }
}
void Kruskal(){
    for(int i=1;i<=n;++i)father[i]=i;
    lowdist=0.0;
    sort(es,es+k,cmp);
    for(int i=0;i<k;++i)
        unite_father(es[i].u,es[i].v,es[i].dist);
}
int main()
{
    while(cin>>n){
        for(int i=1;i<=n;++i)
            cin>>point[i].first>>point[i].second;
        k=0;
        for(int i=1;i<=n;++i){
            for(int j=1;j<=n;++j){
                es[k].u=i;es[k].v=j;
                es[k++].dist=vecx(point[j].first,point[j].second,point[i].first,point[i].second);
            }
        }
        Kruskal();
        cout<<setiosflags(ios::fixed)<<setprecision(2)<<lowdist<<endl;
    }
    return 0;
}