HDU 1162 Eddy's picture （最小生成树）

Eddy's picture

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

Total Submission(s): 8816 Accepted Submission(s): 4485

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

题意：

给N个点求这N个点全联通的最短距离

（将点转化为实际的数1—2—3—N的之间的距离最短的，求出点点之间的距离，然后根据Kruskall算法求即可）

（主要在于点的转化）

for(int i=1; i<=n; i++)
{
for(int j=i+1; j<=n; j++)
{
s[k].x=i;///转化为1——2，1——3···1——n、2——3··2——n·········n-n
s[k].y=j;///
s[k++].d=sqrt(fabs((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])));///距离
}
}

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxx=105;///N>0&&N<=100;
int par[maxx];///存放结点
int n;
int k;///存放边的个数（两两之间）

struct point
{
int x;///x为转化后的起点
int y;///y转化后的终点
double d; ///存放x-y的距离
} s[5005];
int cmp(point a,point b)
{
return a.d<b.d;///按照距离排序（从小到大）
}
///初始化
void init()
{
for(int i=1; i<=n; i++)
par[i]=i;
}
///寻找根节点（路径压缩）
int find(int x)
{
if(par[x]==x)
return x;
else
return par[x]=find(par[x]);
}
///合并
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x!=y)
{
par[x]=y;

}
else
return ;
}
///Kruskall核心算法
void Kruskall()
{

int a,b;
double ll;
double sum=0;
for(int i=1; i<=k; i++)
{
a=s[i].x;
b=s[i].y;
ll=s[i].d;
if(find(a)!=find(b))
{
sum+=ll;
unite(a,b);
}

}
printf("%.2lf\n",sum);

}

int main()
{
while(~scanf("%d",&n))
{
double x[maxx],y[maxx];
for(int i=1; i<=n; i++)
{
scanf("%lf%lf",&x[i],&y[i]);
}
k=1;
for(int i=1; i<=n; i++)
{
for(int j=i+1; j<=n; j++)
{
s[k].x=i;///转化为1——2，1——3···1——n、2——3··2——n·········n-n
s[k].y=j;///
s[k++].d=sqrt(fabs((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])));///距离
}
}
sort(s+1,s+k+1,cmp);
init();
Kruskall();

}
return 0;
}