uva 10034 - Freckles

In an episode of the Dick Van Dyke show, little Richie connects the freckles on his Dad’s back to form a picture of the Liberty Bell. Alas, one of the freckles turns out to be a scar, so his Ripley’s engagement falls through.

Consider Dick’s back to be a plane with freckles at various (x,y) locations. Your job is to tell Richie how to connect the dots so as to minimize the amount of ink used. Richie connects the dots by drawing straight lines between pairs, possibly lifting the pen between lines. When Richie is done there must be a sequence of connected lines from any freckle to any other freckle.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line contains 0 < n <= 100, the number of freckles on Dick’s back. For each freckle, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the freckle.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the freckles.

Sample Input

1

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

最小连通子图问题，有一点比较麻烦，就是需要计算两点之间的距离

声明一个结构体，写一个成员函数计算两点之间距离，每次调用这个成员函数就行了

使用prim算法解决就行了

输出的时候两组数据之间要加一个空行

#include <iostream>
#include <cstdio>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
const int N = 105;
double coor[N][2];
int n, father[N];
struct Node
{
    int x, y;
    double len;
    friend bool operator <(const Node & a, const Node &b)
    {
        return a.len < b.len;
    }
    void fun()
    {
        int a = coor[x][0] - coor[y][0], b = coor[x][1] - coor[y][1];
        len = sqrt(a * a + b * b +0.0);
    }
}node[5100];

int find(int x)
{
    if (x != father[x])
    {
        father[x] = find(father[x]);
    }
    return father[x];
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int t, i, j, k, a, b;
    double ans;
    cin >> t;
    while(t--)
    {
        ans = 0;
        scanf("%d", &n);
        for (i = 0; i < n; i++)
        {
            scanf("%lf%lf", &coor[i][0], &coor[i][1]);
            father[i] = i;
        }
        for (i = 0, k = 0; i < n; i++)
        {
            for (j = 0; j < i; j++)
            {
                node[k].x = i;
                node[k].y = j;
                node[k].fun();
                k++;
            }
        }
        sort(node, node + k);
        for(i = 0; i < k ; i++)
        {
            a = find(node[i].x);
            b = find(node[i].y);
            if (a == b)
            {
                continue;
            }
            ans += node[i].len;
            father[a] = b;
        }
        printf("%.2lf\n", ans);
        if (t)
        {
            cout << endl;
        }
    }

    return 0;
}