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：给出一些点，求所有点连起来的最短路径，算是最小生成树问题，本题使用Kruskal算法
代码：
#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
#define INF 0x3f3f3f3f
#define MAX 101
double c[MAX][MAX];
double dis[MAX],visit[MAX];
typedef struct point
{
double x;
double y;
}Point;
//求两点之间的距离
double getDistance(Point a,Point b)
{
double len;
len=sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y));
return len;
}
int main()
{
Point p[105];
int i,j,k,n;
double sum,l;
while(cin >> n)
{ //输出坐标点
for(i=1;i<=n;i++)
cin >> p[i].x >> p[i].y;
//初始化所有点之间的距离
for(i=0;i<=n;i++)
for(j=0;j<=n;j++)
{
if(i==j)c[i][j]=0;
else c[i][j]=INF;
}
//初始化各个点之间的距离
for(i=1;i<n;i++)
for(j=i+1;j<=n;j++)
{
l=getDistance(p[i],p[j]);
c[i][j]=c[j][i]=l;
}
//初始化起始点到各点的距离
for(i=1;i<=n;i++)
{
dis[i]=c[1][i];
visit[i]=0;
}
//标记1点已经查过
visit[1]=1;
sum=0;
for(i=1;i<n;i++)
{
double minlen=1.0*INF;
//找到与i点到其他点k最短的路径
for(j=1;j<=n;j++)
if(visit[j]==0 && dis[j]<minlen)
{
minlen=dis[j];
k=j;
}
sum+=minlen;
//将k点标记查过
visit[k]=1;
for(j=1;j<=n;j++)
if(visit[j]==0 && dis[j]>c[j][k])dis[j]=c[j][k];
}
// cout << sum << endl;
printf("%.2f\n",sum);
}
return 0;
}
Eddy's picture