Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10085 Accepted Submission(s): 5094

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

1.0 1.0

2.0 2.0

2.0 4.0

Sample Output

3.41

题目大意：给出一些点的坐标，然后连线使得所有顶点都能够连通，并且线的长度最短。

解题思路：根据题意，可以选用以边为主导的Kruskal算法计算。

AC代码：

19975116 2017-03-03 08:53:42 Accepted 1162 0MS 1524K 1304 B G++ 
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>

using namespace std;

struct point
{
    int x,y;
    double l;
}p[11000];
int parent[110],n;
double x[110],y[110];
bool cmp(point a, point b)
{
    return a.l < b.l;
}
int find(int x)
{
    int s,tmp;
    for (s = x; parent[s] >= 0; s = parent[s]);
    while (s != x)
    {
        tmp = parent[x];
        parent[x] = s;
        x = tmp;
    }
    return s;
}
void Union(int A, int B)
{
    int a = find(A), b = find(B);
    int tmp = parent[a]+parent[b];
    if (parent[a] < parent[b])
    {
        parent[b] = a;
        parent[a] = tmp;
    }
    else
    {
        parent[a] = b;
        parent[b] = tmp;
    }
} 
void kruskal(int k)
{
    double sum = 0;
    int u,v,i,j = 0;
    memset(parent,-1,sizeof(parent));
    for (i = 0; i <= k; i ++)
    {
        u = p[i].x;    v = p[i].y;
        if (find(u) != find(v))
        {
            sum += p[i].l;
            Union(u,v);
            j ++;
        }
        if (j == n-1) break;
    }
    printf("%.2lf
",sum);
}
int main ()
{
    int i,j,k;
    while (scanf("%d",&n)!=EOF)
    {
        k = -1;
        for (i = 1; i <= n; i ++)
            scanf("%lf%lf",x+i,y+i);
        for (i = 1; i < n; i ++)
        for (j = 1+i; j <= n; j ++)
        {
            p[++ k].x = i;
            p[k].y = j;
            p[k].l = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
        }
        sort(p,p+k+1,cmp);
        kruskal(k);
    }
    return 0;
}

算法理解： http://www.cnblogs.com/yoke/p/6506492.html